周赛题 （2015年浙江省赛）

ZOJ - 3869

Ace of Aces

Time Limit: 2000MS Memory Limit: 65536KB 64bit IO Format: %lld & %llu

SubmitStatus

Description

There is a mysterious organization called Time-Space Administrative Bureau (TSAB) in the deep universe that we humans have not discovered yet. This year, the TSAB decided to elect an outstanding member from its elite troops. The elected guy will be honored with the title of "Ace of Aces".

After voting, the TSAB received N valid tickets. On each ticket, there is a numberA_i denoting the ID of a candidate. The candidate with the most tickets nominated will be elected as the "Ace of Aces". If there are two or more candidates have the same number of nominations, no one will win.

Please write program to help TSAB determine who will be the "Ace of Aces".

< h4< body="">

Input

There are multiple test cases. The first line of input contains an integer T indicating the number of test cases. For each test case:

The first line contains an integer N (1 <= N <= 1000). The next line containsN integersA_i (1 <= A_i <= 1000).

< h4< body="">

Output

For each test case, output the ID of the candidate who will be honored with "Ace of Aces". If no one win the election, output "Nobody" (without quotes) instead.

< h4< body="">

Sample Input

352 2 2 1 151 1 2 2 31998

< h4< body="">

Sample Output

2Nobody998

Hint

Source

The 12th Zhejiang Provincial Collegiate Programming Contest

//题意：

找出出现次数最多的数，如果有多个输出Nobody

#include<stdio.h>#include<string.h>#include<algorithm>#include<iostream>using namespace std;const int MAXN = 1010;int num[MAXN];int p[MAXN];int main(){int T, N;scanf("%d", &T);while(T--){memset(num, 0, sizeof(num));memset(p, 0, sizeof(p));scanf("%d", &N);int x;for(int i = 0; i < N; i++){scanf("%d", &x);num[x]++;}int ans = 0, temp = 0;for(int i = 0; i < MAXN; i++){if(num[i] >= temp){temp = num[i];ans = i;p[temp]++;}}if(p[num[ans]] > 1){puts("Nobody");}elseprintf("%d\n", ans);}return 0;}

ZOJ - 3870

Team Formation

Time Limit:                                                        3000MS                       Memory Limit:                                                        131072KB                       64bit IO Format:                            %lld & %llu                      

SubmitStatus

Description

For an upcoming programming contest, Edward, the headmaster of Marjar University, is forming a two-man team fromN students of his university.

Edward knows the skill level of each student. He has found that if two students with skill levelA andB form a team, the skill level of the team will be A ⊕ B, where ⊕ means bitwise exclusive or. A team will play well if and only if the skill level of the team is greater than the skill level of each team member (i.e.A ⊕B > max{A, B}).

Edward wants to form a team that will play well in the contest. Please tell him the possible number of such teams. Two teams are considered different if there is at least one different team member.

< h4< body="">

Input

There are multiple test cases. The first line of input contains an integer T indicating the number of test cases. For each test case:

The first line contains an integer N (2 <= N <= 100000), which indicates the number of student. The next line containsN positive integers separated by spaces. Thei^th integer denotes the skill level ofi^th student. Every integer will not exceed 10⁹.

< h4< body="">

Output

For each case, print the answer in one line.

< h4< body="">

Sample Input

231 2 351 2 3 4 5

< h4< body="">

Sample Output

16

Hint

Source

The 12th Zhejiang Provincial Collegiate Programming Contest

//题意：输入一个n，接下来输入n个数。

现在有n个人，n个数代表他们的分数，现在要求从中选出来两个人，并且要求这两个人的异或值要大于这两个数，问有几种情况。

//思路：

因为是异或值所以只用使得两个数的高位上对应的位置出现0和1就可以得到比他俩大的数了。

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<cstring>#include<algorithm>const int MAXN = 100010;int a[MAXN], num[100];typedef long long LL;int geth(int x){int i;for(i = 0;;i++){if((1 << i) > x)break;}return i;}int main(){int T, N;scanf("%d", &T);while(T--){memset(num, 0, sizeof(num));scanf("%d", &N);for(int i = 0; i < N; i++){scanf("%d", a + i);num[geth(a[i])]++;}LL ans = 0;for(int i = 0; i < N; i++){if(a[i] > 0 && a[i] % 2 == 0)ans += num[0];for(int j = 1; (1 << (j - 1)) < a[i]; j++){if((a[i] ^ (1 << (j - 1))) > a[i])ans += num[j];}}printf("%lld\n", ans);}return 0;}

ZOJ - 3872

Beauty of Array

Time Limit:                                                        2000MS                       Memory Limit: 65536KB 64bit IO Format:                            %lld & %llu                      

SubmitStatus

Description

Edward has an array A with N integers. He defines the beauty of an array as the summation of all distinct integers in the array. Now Edward wants to know the summation of the beauty of all contiguous subarray of the arrayA.

< h4< body="">

Input

There are multiple test cases. The first line of input contains an integer T indicating the number of test cases. For each test case:

The first line contains an integer N (1 <= N <= 100000), which indicates the size of the array. The next line containsN positive integers separated by spaces. Every integer is no larger than 1000000.

< h4< body="">

Output

For each case, print the answer in one line.

< h4< body="">

Sample Input

351 2 3 4 532 3 342 3 3 2

< h4< body="">

Sample Output

1052138

Hint

Source

The 12th Zhejiang Provincial Collegiate Programming Contest

//题意：一个序列的漂亮值的定义为：这个序列中所有不重复的数的和。

给你一个含有n个数的序列，让你找出这个序列的所有连续子序列的漂亮值的和。

//思路：

是一道挺好的题，比赛是队友花了好长时间终于把它A了。

因为n的值很大，所以不可能用两重for循环（肯定会TML），所以就想着用一重for实现。

用sum[i]数组存放前i个数的所有可能的组合的情况的和，用pre[x]数组标记x最后一次出现的位置，通过它来计算x会被累加几次，有了这个思路就可以了。

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<cstring>#include<algorithm>#include<set>#include<vector>using namespace std;long long sum[100000+100],pre[100000+100];int main(){int t;scanf("%d",&t);while(t--){int n;scanf("%d",&n);memset(sum,0,sizeof(sum));memset(pre,0,sizeof(pre));long long ans=0;for(long long i=1;i<=n;i++){long long x;scanf("%lld",&x);sum[i]=sum[i-1]+(i-pre[x])*x;pre[x]=i;ans+=sum[i];}printf("%lld\n",ans);}return 0;}

ZOJ - 3875

Lunch Time

Time Limit:                                                        2000MS                       Memory Limit: 65536KB 64bit IO Format:                            %lld & %llu                      

SubmitStatus

Description

The 999th Zhejiang Provincial Collegiate Programming Contest will be held in Marjar University. The canteen of Marjar University is making preparations for this grand competition. The canteen provides a lunch set of three types: appetizer, main course and dessert. Each type has several dishes with different prices for choosing.

Edward is the headmaster of Marjar University. One day, to inspect the quality of dishes, he go to the canteen and decides to choose amedian set for his lunch. That means he must choose one dish from each of appetizers, main courses and desserts. Each chosen dish should at the median price among all dishes of the same type.

For example, if there are five dessert dishes selling at the price of 2, 3, 5, 10, 30, Edward should choose the dish with price 5 as his dessert since its price is located at the median place of the dessert type. If the number of dishes of a type is even, Edward will choose the dish which is more expensive among the two medians.

You are given the list of all dishes, please write a program to help Edward decide which dishes he should choose.

< h4< body="">

Input

There are multiple test cases. The first line of input contains an integer T indicating the number of test cases. For each test case:

The first line contains three integers S, M and D (1 <=S,M, D <= 100), which means that there are S dishes of appetizer,M dishes of main course and D dishes of dessert.

Then followed by three parts. The first part contains S lines, the second and the last part containsM andD lines respectively. In each line of the three parts, there is a string and an integer indicating the name and the price of a dish. The name of dishes will only consist of non-whitespace characters with no more than 50 characters. The price of dishes are non-negative integers less than or equal to 1000. All dish names will be distinct.

< h4< body="">

Output

For each test case, output the total price of the median set, together with the names of appetizer, main course and dessert, separated by a single space.

< h4< body="">

Sample Input

21 3 2Fresh_Cucumber 4Chow_Mein 5Rice_Served_with_Duck_Leg 12Fried_Vermicelli 7Steamed_Dumpling 3Steamed_Stuffed_Bun 42 3 1Stir-fried_Loofah_with_Dried_Bamboo_Shoot 33West_Lake_Water_Shield_Soup 36DongPo's_Braised_Pork 54West_Lake_Fish_in_Vinegar 48Longjing_Shrimp 188DongPo's_Crisp 18

< h4< body="">

Sample Output

15 Fresh_Cucumber Fried_Vermicelli Steamed_Stuffed_Bun108 West_Lake_Water_Shield_Soup DongPo's_Braised_Pork DongPo's_Crisp

Hint

Source

The 12th Zhejiang Provincial Collegiate Programming Contest

//题意：

输入三个数，分别代表三个等级的东西有多少

然后按顺序输入每个东西的名字与质量数字，然后对于每个等级选出其质量为中位数的加起来，最后输出

//思路：水题，直接模拟

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;struct node{char s[101];int val;}A[200],B[200],C[200];bool cmp(node s1,node s2){return s1.val<s2.val;}int main(){int a,b,c;int t;scanf("%d",&t);while(t--){scanf("%d%d%d",&a,&b,&c);for(int i=0;i<a;i++)scanf("%s%d",&A[i].s,&A[i].val);for(int i=0;i<b;i++)scanf("%s%d",&B[i].s,&B[i].val);for(int i=0;i<c;i++)scanf("%s%d",&C[i].s,&C[i].val);sort(A,A+a,cmp);sort(B,B+b,cmp);sort(C,C+c,cmp);int ans=0;ans=A[a/2].val+B[b/2].val+C[c/2].val;printf("%d %s %s %s\n",ans,A[a/2].s,B[b/2].s,C[c/2].s);}return 0;}

ZOJ - 3876

May Day Holiday

Time Limit:                                                        2000MS                       Memory Limit: 65536KB 64bit IO Format:                            %lld & %llu                      

SubmitStatus

Description

As a university advocating self-learning and work-rest balance, Marjar University has so many days of rest, including holidays and weekends. Each weekend, which consists of Saturday and Sunday, is a rest time in the Marjar University.

The May Day, also known as International Workers' Day or International Labour Day, falls on May 1st. In Marjar University, the May Day holiday is a five-day vacation from May 1st to May 5th. Due to Saturday or Sunday may be adjacent to the May Day holiday, the continuous vacation may be as long as nine days in reality. For example, the May Day in 2015 is Friday so the continuous vacation is only 5 days (May 1st to May 5th). And the May Day in 2016 is Sunday so the continuous vacation is 6 days (April 30th to May 5th). In 2017, the May Day is Monday so the vacation is 9 days (April 29th to May 7th). How excited!

Edward, the headmaster of Marjar University, is very curious how long is the continuous vacation containing May Day in different years. Can you help him?

Input

There are multiple test cases. The first line of input contains an integer T indicating the number of test cases. For each test case, there is an integery (1928 <=y <= 9999) in one line, indicating the year of Edward's query.

Output

For each case, print the number of days of the continuous vacation in that year.

Sample Input

3201520162017

Output

569

Input

Output

Sample Input

Sample Output

Hint

Source

The 12th Zhejiang Provincial Collegiate Programming Contest

//题意：

输入一个数n，表示一个年份，每年的五一劳动节都会放假5天，但是若赶得的巧的话（连着周六周日一块），可能会放6天或者9天，问这一年五一会连着放几天假？

//思路：

通过找规律可以知道，若五一这一天是周一，那么，这一年可能放9天（连着前一个星期的周六周日和下一个星期的周六周日），若五一是周二或者是周日会放6天（递推和前面一样），其他的是放5天，知道这个规律，就可以做这个题了。

根据样例可知2015年五一是周五，2016是周日（遇到闰年，它的五一的星期会与上一年相差两天，平年的相差一天），2017是周一，所以根据这个可以打表。

#include<stdio.h>#include<string.h>#include<algorithm>#include<iostream>using namespace std;int a[10010];int vis[10010];int init(){a[2015]=5;for(int i=2016;i<=10000;i++){if(i%4==0&&i%100!=0||i%400==0)a[i]=(a[i-1]+2)%7;elsea[i]=(a[i-1]+1)%7;}int k=5;for(int i=2015;i>=1928;i--){if(i%4==0&&i%100!=0||i%400==0){if(k>=2)a[i-1]=k-2;else{k=k+7;a[i-1]=k-2;}k-=2;}else{if(k>=1)a[i-1]=k-1;else{k+=7;a[i-1]=k-1;}k--;}}}int main(){int t,n;init();scanf("%d",&t);while(t--){scanf("%d",&n);if(a[n]==1)printf("9\n");else if(a[n]==0||a[n]==2)printf("6\n");elseprintf("5\n");}return 0;}

ZOJ - 3878

Convert QWERTY to Dvorak

Time Limit:                                                        2000MS                       Memory Limit: 65536KB 64bit IO Format:                            %lld & %llu                      

SubmitStatus

Description

Edward, a poor copy typist, is a user of the Dvorak Layout. But now he has only a QWERTY Keyboard with a brokenCaps Lock key, so Edward never presses the brokenCaps Lock key. Luckily, all the other keys on the QWERTY keyboard work well. Every day, he has a lot of documents to type. Thus he needs a converter to translate QWERTY into Dvorak. Can you help him?

The QWERTY Layout and the Dvorak Layout are in the following:

The QWERTY Layout

---

The Dvorak Layout

---

< h4< body="">

Input

A QWERTY document Edward typed. The document has no more than 100 kibibytes. And there are no invalid characters in the document.

< h4< body="">

Output

The Dvorak document.

< h4< body="">

Sample Input

Jgw Gqm Andpw a H.soav Patsfk f;doeNfk Gq.d slpt a X,dokt vdtnsaoheKjd yspps,glu pgld; aod yso kd;kgluZ1234567890`~!@#$%^&*()}"']_+-=ZQqWEwe{[\|ANIHDYf.,bt/ABCDEFuvwxyz

< h4< body="">

Sample Output

Hi, I'm Abel, a Dvorak Layout user.But I've only a Qwerty keyboard.The following lines are for testing:1234567890`~!@#$%^&*()+_-={}[]:"'<>,.?/\|ABCDEFuvwxyzAXJE>Ugk,qf;

Hint

Source

The 12th Zhejiang Provincial Collegiate Programming Contest

//题意：

小明的键盘和我们正常的键盘不同，如上图对应的情况，现在说输入一个字符串，让你输出小明的键盘输出的字符串。

//思路：

水题，直接模拟，但是很坑，题中要求的是输入不超过100K（1K=1024个字符），这块没看清，以为是100个字符，所以就WA了一遍。

#include<stdio.h>#include<string.h>#include<algorithm>#include<iostream>using namespace std;char s[10000010];char c[10000010];int main(){while(gets(s)!=NULL){memset(c,'\0',sizeof(c));int len=strlen(s);for(int i=0;i<len;i++){if(s[i]=='_') c[i]='{';else if(s[i]=='-') c[i]='[';else if(s[i]=='+') c[i]='}';else if(s[i]=='=') c[i]=']';else if(s[i]=='Q') c[i]='"';else if(s[i]=='q') c[i]=39;else if(s[i]=='W') c[i]='<';else if(s[i]=='w') c[i]=',';else if(s[i]=='E') c[i]='>';else if(s[i]=='e') c[i]='.';else if(s[i]=='R') c[i]='P';else if(s[i]=='r') c[i]='p';else if(s[i]=='T') c[i]='Y';else if(s[i]=='t') c[i]='y';else if(s[i]=='Y') c[i]='F';else if(s[i]=='y') c[i]='f';else if(s[i]=='U') c[i]='G';else if(s[i]=='u') c[i]='g';else if(s[i]=='I') c[i]='C';else if(s[i]=='i') c[i]='c';else if(s[i]=='O') c[i]='R';else if(s[i]=='o') c[i]='r';else if(s[i]=='P') c[i]='L';else if(s[i]=='p') c[i]='l';else if(s[i]=='{') c[i]='?';else if(s[i]=='[') c[i]='/';else if(s[i]=='}') c[i]='+';else if(s[i]==']') c[i]='=';else if(s[i]=='S') c[i]='O';else if(s[i]=='s') c[i]='o';else if(s[i]=='D') c[i]='E';else if(s[i]=='d') c[i]='e';else if(s[i]=='F') c[i]='U';else if(s[i]=='f') c[i]='u';else if(s[i]=='G') c[i]='I';else if(s[i]=='g') c[i]='i';else if(s[i]=='H') c[i]='D';else if(s[i]=='h') c[i]='d';else if(s[i]=='J') c[i]='H';else if(s[i]=='j') c[i]='h';else if(s[i]=='K') c[i]='T';else if(s[i]=='k') c[i]='t';else if(s[i]=='L') c[i]='N';else if(s[i]=='l') c[i]='n';else if(s[i]==':') c[i]='S';else if(s[i]==';') c[i]='s';else if(s[i]=='"') c[i]='_';else if(s[i]==39) c[i]='-';else if(s[i]=='Z') c[i]=':';else if(s[i]=='z') c[i]=';';else if(s[i]=='X') c[i]='Q';else if(s[i]=='x') c[i]='q';else if(s[i]=='C') c[i]='J';else if(s[i]=='c') c[i]='j';else if(s[i]=='V') c[i]='K';else if(s[i]=='v') c[i]='k';else if(s[i]=='B') c[i]='X';else if(s[i]=='b') c[i]='x';else if(s[i]=='N') c[i]='B';else if(s[i]=='n') c[i]='b';else if(s[i]=='<') c[i]='W';else if(s[i]==',') c[i]='w';else if(s[i]=='>') c[i]='V';else if(s[i]=='.') c[i]='v';else if(s[i]=='?') c[i]='Z';else if(s[i]=='/') c[i]='z';else c[i]=s[i];}puts(c);}return 0;}

ZOJ - 3879

Capture the Flag

Time Limit:                                                        2000MS                       Memory Limit: 65536KB 64bit IO Format:                            %lld & %llu                      

SubmitStatus

Description

In computer security, Capture the Flag (CTF) is a computer security competition. CTF contests are usually designed to serve as an educational exercise to give participants experience in securing a machine, as well as conducting and reacting to the sort of attacks found in the real world. Reverse-engineering, network sniffing, protocol analysis, system administration, programming, and cryptanalysis are all skills which have been required by prior CTF contests at DEF CON. There are two main styles of capture the flag competitions: attack/defense and jeopardy.

In an attack/defense style competition, each team is given a machine (or a small network) to defend on an isolated network. Teams are scored on both their success in defending their assigned machine and on their success in attacking other team's machines. Depending on the nature of the particular CTF game, teams may either be attempting to take an opponent's flag from their machine or teams may be attempting to plant their own flag on their opponent's machine.

Recently, an attack/defense style competition called MCTF held by Marjar University is coming, and there areN teams which participate in the competition. In the beginning, each team hasS points as initial score; during the competition, there are some checkpoints which will renew scores for all teams. The rules of the competition are as follows:

• If a team has been attacked for a service P, they will lose N - 1 points. The lost points will be split equally and be added to the team(s) which attacks successfully. For example, there are 4 teams and Team A has been attacked by Team B and Team C, so Team A will lose 3 points, while Team B and Team C each will get 1.5 points.
• If a team cannot maintain their service well, they will lose N - 1 points, which will be split equally too and be added to the team(s) which maintains the service well.

The score will be calculated at the checkpoints and then all attacks will be re-calculated. Edward is the organizer of the competition and he needs to write a program to display the scoreboard so the teams can see their scores instantly. But he doesn't know how to write. Please help him!

< h4< body="">

Input

There are multiple test cases. The first line of input contains an integer T indicating the number of test cases. For each test case:

The first line contains four integers N (2 <= N <= 100) - the number of teams,Q - the number of services (1 <=Q <= 10), S - the initial points (0 <=S <= 100000) andC - the number of checkpoints (1 <= C <= 100).

For each checkpoint, there are several parts:

• The first line contains an integer A - the number of the successful attacks. ThenA lines follow and each line contains a message:

  [The No. of the attacker] [The No. of the defender] [The No. of the service]

  For example, "1 2 3" means the 1^st team attacks the 2^nd team in service 3 successfully. The No. of teams and services are indexed from 1. You should notice that duplicate messages are invalid because of the rules. Just ignore them.
• Then there are Q lines and each line contains N integers. Thej^th number of thei^th line indicating thej^th team's maintaining status of thei^th service, where 1 means well and 0 means not well.
• Finally there is an integer U (0 <= U <= 100), which describing the number of the queries. The following line containsU integers, which means Edward wants to know the score and the ranking of these teams.

< h4< body="">

Output

For each query L, output the score and the ranking of the L^th team. The relative error or absolute error of the score should be less than 10^-5. The team with higher score gets higher rank; the teams with the same scores should have the same rank. It is guaranteed that the scores of any two teams are either the same or with a difference greater than 10^-5.

< h4< body="">

Sample Input

14 2 2500 501 1 1 11 1 1 141 2 3 421 2 13 2 11 1 1 11 1 1 141 2 3 411 2 21 1 1 11 1 1 041 2 3 400 0 0 00 0 0 041 2 3 401 1 1 11 1 1 121 4

< h4< body="">

Sample Output

2500.00000000 12500.00000000 12500.00000000 12500.00000000 12501.50000000 12497.00000000 42501.50000000 12500.00000000 32505.50000000 12495.00000000 42502.50000000 22497.00000000 32499.50000000 12489.00000000 42496.50000000 22491.00000000 32499.50000000 12491.00000000 3

Hint

For C++ users, kindly use scanf to avoid TLE for huge inputs.

Hint

Source

The 12th Zhejiang Provincial Collegiate Programming Contest

//题意：有N个人相互打架，掠夺财产

前提：

1、在一个事件中，如果人x在某个水晶塔旁边被k个人打了，那么人x的钱数会将少（N-1）元，这（N-1）元会被打他的k个人平分。

2、如果人x的某个水晶塔坏了，那么他就得花（N-1）元请这个水晶塔好着的人来修复，那么这（N-1）元就会被他们平分。

先输入四个数N,Q,S,C，分别表示N个人参加一场搏斗，有Q个水晶塔（服务站），每个队伍刚开始时都有S元钱，接下来是C个事件。

每个事件第一行先输入一个A，表示发生了A次战斗，接下来A行，每行输入上个数x,y,z,表示在第z个水晶塔旁边x击打了y。（要忽略重复的）

接下来有Q行输入，每行输入有N个数，表示第j个人的第i个水晶塔是否是好的（1表示好的，0表示坏的）。

再接下来输入一个u，表示有u个询问，接下来输入u个数，对于每个数，表示询问这个人现在的钱数和排名，并输出。

//思路：

题意理解清了，这就是个水题了，直接模拟

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<cstring>#include<algorithm>#include<set>#include<vector>using namespace std;typedef long long LL;struct Node{double sc;int k, rat;friend bool operator < (Node a, Node b){if(a.sc >= b.sc) return true;else return false;}};Node dt[10010], ans[10010];int vis[110][110][110];vector<int>vec[110][110];struct Node1{int b, c;};Node1 vv[100010];int sj[1010];int main(){int T, N, Q, S, C;scanf("%d", &T);while(T--){scanf("%d%d%d%d", &N, &Q, &S, &C);for(int i = 1; i <= N; i++)dt[i].k = i, dt[i].sc = S;while(C--){memset(vis, 0, sizeof(vis));for(int i = 0; i < 110; i++)for(int j = 0; j < 110; j++)vec[i][j].clear();int tp = 0;int A;scanf("%d", &A);for(int j = 0; j < A; j++){int a, b, c;scanf("%d%d%d", &a, &b, &c);if(!vis[a][b][c]){vis[a][b][c] = 1;if(vec[b][c].size() == 0){vv[tp].b = b;vv[tp].c = c;tp++;}vec[b][c].push_back(a);}}for(int j = 0; j < tp; j++){dt[vv[j].b].sc -= (N - 1);for(int i = 0; i < vec[vv[j].b][vv[j].c].size(); i++){int a = vec[vv[j].b][vv[j].c][i];dt[a].sc += 1.0*(N - 1)/vec[vv[j].b][vv[j].c].size();}}for(int j = 1; j <= Q; j++){int t = 0;for(int i = 1; i <= N; i++){scanf("%d", sj + i);if(sj[i])t++;}for(int i = 1; i <= N; i++){if(sj[i] == 0){dt[i].sc -= (N - 1);for(int k = 1; k <= N; k++){if(sj[k]){dt[k].sc += 1.0*(N - 1)/t;}}}}}for(int i = 0; i <= 110; i++)ans[i] = dt[i];sort(ans + 1, ans + N + 1);ans[1].rat = 1;for(int i = 2; i <= N; i++){if(fabs(ans[i].sc - ans[i - 1].sc) <= 1e-6){ans[i].rat = ans[i - 1].rat;}else ans[i].rat = i;}for(int i = 1; i <= N; i++){dt[ans[i].k].rat = ans[i].rat;}int L;scanf("%d", &L);while(L--){int x;scanf("%d", &x);for(int i = 1; i <= N; i++){if(dt[i].k == x){printf("%lf %d\n", dt[i].sc, dt[i].rat);break;}}}}}return 0;}

ZOJ - 3880

Demacia of the Ancients

Time Limit:                                                        2000MS                       Memory Limit: 65536KB 64bit IO Format:                            %lld & %llu                      

SubmitStatus

Description

There is a popular multiplayer online battle arena game called Demacia of the Ancients. There are lots of professional teams playing this game. A team will be approved as LevelK if there are exactK team members whose match making ranking (MMR) is strictly greater than 6000.

You are given a list of teams. Please calculate the level of each team.

< h4< body="">

Input

There are multiple test cases. The first line of input contains an integer T indicating the number of test cases. For each test case:

The first line contains an integer N (1 <= N <= 10) indicating the number of team members.

The second line contains N integers representing the MMR of each team member. All MMRs are non-negative integers less than or equal to 9999.

< h4< body="">

Output

For each test case, output the level of the given team.

< h4< body="">

Sample Input

357986 6984 6645 6200 615057401 7377 6900 6000 43003800 600 200

< h4< body="">

Sample Output

530

Hint

Source

The 12th Zhejiang Provincial Collegiate Programming Contest

题意：

给你n个数，问你这n个数中有几个数大于6000.

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<cstring>#include<algorithm>int main(){int t;scanf("%d",&t);while(t--){int n;int ans=0;int x;scanf("%d",&n);for(int i=0;i<n;i++){scanf("%d",&x);if(x>6000)ans++;}printf("%d\n",ans);}return 0;}