2011 Heilongjiang collegiate programming contest题解报告

题目：

UESTC 924~933

A.水题

B.水题

C.模拟， 也没什么坑点.

D 

Dart game

Problem:696

Time Limit:1000ms

Memory Limit:65536K

Description

　　　Darts originated in Australia. Australia's aborigines initially for hunting and hit the enemy's weapon. 　　　A game of darts in which the players attempt to score points by throwing the darts at a target.
Figure：DART BOARD　　　Darts movement rules of the game is very simple, the target is 1-20 points and the central circle is 50 small zoning, edge is 25 division, the rough circle line of fan-shaped covered area is 1-20 points and three times the corresponding division. This game is generally played by two people but can be played by teams. Each player starts with N points. The goal for each player is to reach zero by subtracting the amount they score from the amount they had left,but final throwing must be double division. And the first to reduce his/her score to zero wins.　　　So the task is : 　　　Given a dart scores N that a player starts with, you are required to calculate how many different ways to reach zero. One is different way to another means at least one dart hits different division.Ways which have different orders and same divisions are the same way. For example,if N=4,there are 4 different ways reach to zero:the first is double 2,the second is 2 and double 1, the third is twice of double 1,the fourth is twice of 1 and double 1 .　　　The answer may be very large,you have to module it by 2011. 

Input

The input contains several test cases.   Each test case contains an integer N ( 0 < N ≤ 1001 ) in a line.   N=0 means end of input and need not to process.

Output

For each test case, output how many different ways to reach zero.

Sample Input

543210

Sample Output

64110

Hint

　　5=1+1×2+1×2　　5=1+1+1+1×2　　5=3+1×2　　5=1×3+1×2　　5=1+2+1×2　　5=1+2×2　　4=2×2　　4=1+1+1×2　　4=2+1×2　　4=1×2+1×2　　3=1+1×2　　2=1×2　　1:no way

题目大意：（不是我说，题意真的是难懂）

现在有1-20这些分值，每个数都可以得到i分，或者是i*2分，或者是i*3分。

中间有一个25分，可以得到25分或者是50分。

问你必须最后一次得分为二倍得分的方案数，使得得分加和为N分。

思路：就是多重背包求整数划分方案数，也算是跟包巨学了一招，比较好的点是， 他说最后一次一定是2*x， 那我们就用所有情况的方案数-一个2*x也没有的， 就是至少一个2*x，因为无关顺序，那我们肯定有一个2*x放后面~

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int Mod = 2011;const int maxn = 2e3 + 5;int dp1[maxn], dp2[maxn];void init(){    dp1[0] = dp2[0] = 1;    for(int i = 1; i <= 20; i++)    {        for(int j = i; j < maxn; j++)        {            dp1[j] = (dp1[j-i] + dp1[j])%Mod; //方案数求法            dp2[j] = (dp2[j-i] + dp2[j])%Mod;        }    }    for(int i = 25; i < maxn; i++)            dp1[i] = (dp1[i]+dp1[i-25])%Mod, dp2[i] = (dp2[i]+dp2[i-25])%Mod; //求25的    for(int i = 50; i < maxn; i++)            dp1[i] = (dp1[i]+dp1[i-50])%Mod; //求50的    for(int i = 1; i <= 20; i++)   //求2的倍数作为分子的        for(int j = i*2; j < maxn; j++)              dp1[j] = (dp1[j]+dp1[j-i*2])%Mod;     for(int i = 1; i <= 20; i++)  //求3的倍数        for(int j = i*3; j < maxn; j++)            dp1[j] = (dp1[j]+dp1[j-i*3])%Mod, dp2[j] = (dp2[j]+dp2[j-i*3])%Mod;}int main(){    int n;    init();    while(~scanf("%d", &n), n)    {        printf("%d\n", (dp1[n]-dp2[n]+Mod)%Mod);  //注意分别取模的相减，一定要+MOD再取模，一定！！！    }    return 0;}

E

Similar Word

Time Limit: 3000/1000MS (Java/Others)     Memory Limit: 65535/65535KB (Java/Others)

Submit Status

It was a crummy day for Lur. He failed to pass to the CET-6 (College English Test Band-66). Looking back on how it was in last year gone by, he gradually noticed he had fled too many English Lessons. But he determines to memorize words on his bed ,not in the classroom. You know, it is not that easy to pass the test mainly because the large amount of born words.

Lur is intelligent on games , never English. He cann't learn the similar words by heart. He always choose to select a word to learn from the similar words . For him, two words are similar if and only if one word can equal to the other by multiple cyclic shift(at least 11). For example, car and arc are similar words, while car and rca are also similar words . To save more time to play games.

Lur want to know wether two words are similar words faster, he asks you to write a program to tell him ,can you help him ?

Input

There are multiple test cases. Each case contains two lines. Each line contains a word, WW. You can assume that length(W)≤105(W)≤105 . Ended by EOF.

Output

Output yes in a single line if two words are similar,otherwise you should output no in a single line.

Sample input and output

Sample InputSample Output

cararccarcracarcar

yesnono

Source

2011 Heilongjiang collegiate programming contest

 

 算是水题，把字符串复制一次， 然后kmp匹配就好，注意长度不相同是no.

#include<iostream>#include<cstdio>#include<cstring>using namespace std;const int maxn = 1e6+5;char s[maxn], t[maxn];int Next[maxn], ans;void makeNext(void){    int len = strlen(s);    Next[0] = Next[1] = 0;    for(int i = 1; i < len; i++)    {        int j = Next[i];        while(j && s[i] != s[j]) j = Next[j];        Next[i+1] = s[i]==s[j] ? j+1 : 0;    }}void kmp(void){    int len1 = strlen(s);    int len2 = strlen(t);    int i, j = 0;    for(int i = 0; i < len1; i++)    {        while(j && s[i] != t[j]) j = Next[j];        if(s[i] == t[j]) j++;        if(j == len2) ans++, j = Next[j];    }}int main(void){    while(~scanf(" %s %s", s, t))    {        ans = 0;        int len = strlen(s);        int len2 = strlen(t);        if(len != len2) { puts("no"); continue ; }        bool same = 1;        for(int i = 0; i < len; i++)        {            if(s[i] != t[i])                same = 0;        }        for(int i = 0; i < len; i++)            s[i+len] = s[i];        s[len*2] = 0;        makeNext();        kmp();//        cout << ans << ' ' << same << endl;        if(ans == 2 && same) puts("no");        else if(ans) puts("yes");        else puts("no");    }    return 0;}

F 前缀和+贪心  跟这差不多点击打开链接

G

Lucky Boy

Time Limit: 3000/1000MS (Java/Others)     Memory Limit: 65535/65535KB (Java/Others)

Submit Status

Recently, Lur have a good luck. He is also the cleverest boy in his school as he create the most popular computer game – Lucky Boy. The game is played by two players, aa and bb, in 22d planar .In the game Lucky Boy, there are nn different points on plane, each time one can remove one or multiple co-line points from the plane. The one who can firstly remove more than two points from the plane wins. The one who removes the last point on the plane can also win the game. You may assume that two players are both clever enough that they can always make the best choice. The winner is called Lucky Boy.

Given the n points, can you tell me who will be the Lucky Boy ? Note that player a will always the first one to remove points from the plane.

Input

The first line of each case is an integer n(0<n≤103)n(0<n≤103), following nn lines each contains two integers xx and y(0≤x,y≤108)y(0≤x,y≤108), describing the coordinates of each point. Ended by EOF.

Output

Output a is the lucky boy. in a single line if a win the game, otherwise you should output b is the lucky boy. in a single line.

Sample input and output

Sample InputSample Output

30 01 12 230 01 12 3

a is the lucky boy.b is the lucky boy.

思路： 主要难在判断三点共线， 可以n2判断， 不用n3枚举三个点， 枚举每个点， 把它跟其余n-1点做斜率， 如果有斜率相等的，说明共线， 我用的是doube存在map也可以化简成分数， 存在一个二维map里，这样精度很高~注意共线跟平行不一样，平行且相等，直接向量搞， 三点共线， 就枚举中间点就好了~

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <map>using namespace std;const int maxn = 1e3 + 5;struct node{    int x, y;}a[maxn];int main(){    int n;    while(~scanf("%d", &n))    {        for(int i = 1; i <= n; i++)            scanf("%d%d", &a[i].x, &a[i].y);        int flag = 0;        for(int i = 1; i <= n; i++)        {            if(flag) break;            map<double, int> mp;            for(int j = i+1; j <= n; j++)            {                if(a[j].y == a[i].y && !mp[-1])                    mp[-1]++;                else if(a[j].y == a[i].y && mp[-1])                {                    flag = 1;                    break;                }                else                {                    double tmp = (a[j].x-a[i].x)*1.0 / (a[j].y-a[i].y);                    if(mp[tmp])                    {                        flag = 1;                        break;                    }                    else mp[tmp]++;                }            }        }        if(flag)            puts("a is the lucky boy.");        else        {            if(n%3 == 0)                 puts("b is the lucky boy.");            else                 puts("a is the lucky boy.");        }    }    return 0;}

H

Car race game

Time Limit: 3000/1000MS (Java/Others)     Memory Limit: 65535/65535KB (Java/Others)

Submit Status

Bob is a game programming specialist. In his new car race game, there are some racers(nn means the amount of racers (1≤n≤100000)(1≤n≤100000)) racers star from someplace(xiximeans Starting point coordinate),and they possible have different speed(VV means speed).so it possibly takes place to overtake(include staring the same point ). now he want to calculate the maximal amount of overtaking.

Input

The first line of the input contains an integer nn-determining the number of racers. Next nn lines follow, each line contains two integer XiXi and ViVi.(xixi means the ithith racer's Starting point coordinate, ViVi means the ith racer's speed.0<Xi,Vi<10000000<Xi,Vi<1000000).

Output

For each data set in the input print on a separate line, on the standard output, the integer that represents the maximal amount of overtaking.

Sample input and output

Sample InputSample Output

22 12 252 69 43 14 99 175 56 105 63 109 109 52 2

167

Source

2011 Heilongjiang collegiate programming contest

思路：排序+树状数组，水

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;typedef long long ll;const int maxn = 1e6 + 5;const int maxm = 1e7 + 5;int c[maxm];struct node{    int x, v;}a[maxn];int cmp(node a, node b){    if(a.x == b.x)        return a.v > b.v;    return a.x < b.x;}void update(int k){    while(k < maxm)    {        c[k] += 1;        k += k & -k;    }}ll sum(int k){    ll s = 0;    while(k > 0)    {        s += c[k];        k -= k & -k;    }    return s;}int main(){    int n;    while(~scanf("%d", &n))    {        memset(c, 0, sizeof(c));        int maxx = -1;        for(int i = 1; i <= n; i++)        {            scanf("%d%d", &a[i].x, &a[i].v);            maxx = max(maxx, a[i].v);        }        sort(a+1, a+1+n, cmp);        ll ans = 0;        for(int i = 1; i <= n; i++)        {//            if(a[i-1].x == a[i].x &&  a[i].v == a[i-1].v) ans++;            ans += sum(maxm)-sum(a[i].v);            update(a[i].v);        }        printf("%lld\n", ans);    }    return 0;}

I 实在读不懂那个英文题

队友题解：http://blog.csdn.net/CillyB/article/details/75136792

J

打表找规律，不会就打表，而且这个值可能很大， 肯定有啥规律- -， 注意n*n+n*n肯定是个解，然后打表- -

首先根据p为素数,能想到i*i%p = （i+p）*(i+p)%p的循环节是p.

所以说如果选择数的话 一定在这个循环节里面,

然后能想到i∗i+j∗j=p那么结果就是p 如果找不到 则只能是p∗p+p∗p

然后想到枚举平方数找存不存在i∗i+j∗j=p ,复杂度为O(sqrt(n)) 

#include <bits/stdc++.h>    typedef long long int LL;    using namespace std;    map<LL,LL >mmp;    int main() {      LL n; mmp.clear();      while(~scanf("%lld",&n)) {          if(n%4==3) printf("%lld\n",n*n*2);          else       printf("%d\n",n);      }      return 0;  }