Codeforces Round #306 (Div. 2)

最近是怎么了，老是掉分，sad = =.......

明明都会做，但一到比赛时，就没有想法了。。。估计是太困了吧。。。

A. Two Substrings

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given string s. Your task is to determine if the given string s contains two non-overlapping substrings "AB" and "BA" (the substrings can go in any order).

Input

The only line of input contains a string s of length between 1 and 105 consisting of uppercase Latin letters.

Output

Print "YES" (without the quotes), if string s contains two non-overlapping substrings "AB" and "BA", and "NO" otherwise.

Sample test(s)

input

ABA

output

NO

input

BACFAB

output

YES

input

AXBYBXA

output

NO

Note

In the first sample test, despite the fact that there are substrings "AB" and "BA", their occurrences overlap, so the answer is "NO".

In the second sample test there are the following occurrences of the substrings: BACFAB.

In the third sample test there is no substring "AB" nor substring "BA".

这道题我看了好久，原因呢？我不知道它说的题目意思是两个字符串“AB"和”BA"就可以了，还是要分别都要两个，事后想想，这个愚蠢的问题。。。然后回过头看，wa~，完了，又要掉分了。。。

想法呢：

就是for4遍，首先第一遍是为了先找出AB有没有，然后for第二遍找出BA，记得把AB所在的位置要标记上；

然后如果上面找到两种字符串的话，那么就不用进行下面的语句了，但是没有找到，那么就要for先找出BA,然后就再去找AB，大致与上面找的方法相同。

#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>#include<map>using namespace std;#define maxn 111111map<string,int> mp;char a[maxn];int mp1[maxn],mp2[maxn];int main(){    scanf("%s",a);    int len=strlen(a);    int i,j,k,l;    bool f1,f2;    f1=f2=false;    for(i=0;i<len;i++){        if(a[i]=='A'&&a[i+1]=='B'&&i!=len-1) {mp1[i]=1; mp1[i+1]=1; f1=true; break;}    }    for(i=0;i<len;i++){        if(a[i]=='B'&&a[i+1]=='A'&&!mp1[i]&&!mp1[i+1]&&i!=len-1) {f2=true; break;}    }    if(f1&&f2) printf("YES\n");    else{        f1=f2=false;        memset(mp1,0,sizeof(mp1));        memset(mp2,0,sizeof(mp2));        for(i=0;i<len;i++){            if(a[i]=='B'&&a[i+1]=='A'&&i!=len-1){                mp1[i]=1; mp1[i+1]=1; f1=true; break;            }        }        for(i=0;i<len;i++){            if(a[i]=='A'&&a[i+1]=='B'&&!mp1[i]&&!mp1[i+1]&&i!=len-1){                f2=true; break;            }        }        if(f1&&f2) printf("YES\n");        else printf("NO\n");    }}

B. Preparing Olympiad

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You have n problems. You have estimated the difficulty of the i-th one as integer ci. Now you want to prepare a problemset for a contest, using some of the problems you've made.

A problemset for the contest must consist of at least two problems. You think that the total difficulty of the problems of the contest must be at least l and at most r. Also, you think that the difference between difficulties of the easiest and the hardest of the chosen problems must be at least x.

Find the number of ways to choose a problemset for the contest.

Input

The first line contains four integers n, l, r, x (1 ≤ n ≤ 15, 1 ≤ l ≤ r ≤ 109, 1 ≤ x ≤ 106) — the number of problems you have, the minimum and maximum value of total difficulty of the problemset and the minimum difference in difficulty between the hardest problem in the pack and the easiest one, respectively.

The second line contains n integers c1, c2, ..., cn (1 ≤ ci ≤ 106) — the difficulty of each problem.

Output

Print the number of ways to choose a suitable problemset for the contest.

Sample test(s)

input

3 5 6 11 2 3

output

2

input

4 40 50 1010 20 30 25

output

2

input

5 25 35 1010 10 20 10 20

output

6

Note

In the first example two sets are suitable, one consisting of the second and third problem, another one consisting of all three problems.

In the second example, two sets of problems are suitable — the set of problems with difficulties 10 and 30 as well as the set of problems with difficulties 20 and 30.

In the third example any set consisting of one problem of difficulty 10 and one problem of difficulty 20 is suitable.

这道题我一开始没有想出来，后来看别人的想法做的，就是一个dfs。

题意就是给你限定了一个范围区间[l,r]； 然后你可以选n门课中的任意门然后要使它们的难度系数之和在这个区间范围内，然后在你所选的课中找出最大值max，最小值min，然后它们两个的差值要大于等于x才行，然后问你总共有几种选课的方法。

#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>using namespace std;#define maxn  22#define inf 99999999int a[maxn],l,r,x,n;int num=0;void dfs(int cur,int sum,int ma,int mi,int cnt){    if(sum>r) return;    if(ma<a[cur]) ma=a[cur];    if(mi>a[cur]) mi=a[cur];    if(sum>=l&&sum<=r&&(ma-mi>=x)&&cnt>=2){        num++;        //return    注意了，这里并不用加return，因为我们这里并不需要返回，因为我们要做的就是让它不撞南墙不回头的去搜索！！！     }    for(int i=cur+1;i<n;i++){        dfs(i,sum+a[i],ma,mi,cnt+1);    }}int main(){    scanf("%d%d%d%d",&n,&l,&r,&x);    for(int i=0;i<n;i++) scanf("%d",&a[i]);    //sort(a,a+n);    num=0;    //这里因为我们是从前往后依次搜过去，所以不用担心前面的情况没有被覆盖到，所以下面写成从i开始搜！！     for(int i=0;i<n;i++){        dfs(i,a[i],-1,inf,1);    }    printf("%d\n",num);}

C. Divisibility by Eight

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a non-negative integer n, its decimal representation consists of at most 100 digits and doesn't contain leading zeroes.

Your task is to determine if it is possible in this case to remove some of the digits (possibly not remove any digit at all) so that the result contains at least one digit, forms a non-negative integer, doesn't have leading zeroes and is divisible by 8. After the removing, it is forbidden to rearrange the digits.

If a solution exists, you should print it.

Input

The single line of the input contains a non-negative integer n. The representation of number n doesn't contain any leading zeroes and its length doesn't exceed 100 digits.

Output

Print "NO" (without quotes), if there is no such way to remove some digits from number n.

Otherwise, print "YES" in the first line and the resulting number after removing digits from number n in the second line. The printed number must be divisible by 8.

If there are multiple possible answers, you may print any of them.

Sample test(s)

input

3454

output

YES344

input

10

output

YES0

input

111111

output

NO

这道题我是找规律的，首先我发现后两位数是有重复的，比如说从0~100与200~300的后两位是重复的，从100~200与300~400的后两位是重复的，所以我们可以对后两位进行判断。

代码有点挫，先贴上来，明天再整理下。

#include<stdio.h>#include<string.h>#include<algorithm>#include<iostream>using namespace std;#define maxn 111char a[maxn];int mp[33];int main(){    scanf("%s",a);    int len=strlen(a);    for(int i=0;i<len;i++){        mp[a[i]-'0']++;    }    if(mp[0]||mp[8]){        puts("YES");        if(mp[0]) {printf("0\n"); return 0;}        if(mp[8]) {printf("8\n"); return 0;}    }    else{        if(!mp[2]&&!mp[4]&&!mp[6]) puts("NO");        else{            if(mp[2]){                bool ff=false;                int tx=0;                int ss1,ss2[maxn]={0},ii=0,pos=0;                for(int i=0;i<len;i++){                    if(a[i]-'0'==3) {ss1=i; break;}                }                for(int i=0;i<len;i++){                    if(a[i]-'0'==2) ss2[ii++]=i;                }                for(int i=0;i<len;i++){                    if((a[i]-'0')%2) {tx=a[i]-'0'; pos=i; ff=true; break;}                }                bool flag=false;                for(int i=0;i<ii;i++){                    if(ss1<ss2[i])  {flag=true; break;}                }                if(mp[3]&&flag) {puts("YES"); printf("32\n");return 0;}                for(int i=0;i<len;i++){                    if(a[i]-'0'==7) {ss1=i; break;}                }                flag=false;                for(int i=0;i<ii;i++){                    if(ss1<ss2[i])  {flag=true; break;}                }                if(mp[7]&&flag) {puts("YES"); printf("72\n");return 0;}                int nn=0,ss3=0;                for(int i=0;i<len;i++){                    if(a[i]-'0'==1&&nn==0) {ss1=i; nn++;}                    if(a[i]-'0'==1&&nn==1) {ss3=i,nn++;break;}                }                flag=false;                for(int i=0;i<ii;i++){                    if(ss1<ss2[i]&&ss3<ss2[i])  {flag=true; break;}                }                if(mp[1]>=2&&flag) {puts("YES"); printf("112\n");return 0;}                for(int i=0;i<len;i++){                    if(a[i]-'0'==5) {ss1=i; break;}                }                flag=false;                for(int i=0;i<ii;i++){                    if(ss1<ss2[i]&&pos<ss2[i]&&pos<ss1)  {flag=true; break;}                }                if(ff&&mp[5]&&flag) {puts("YES"); printf("%d52\n",tx);return 0;}                for(int i=0;i<len;i++){                    if(a[i]-'0'==9) {ss1=i; break;}                }                flag=false;                for(int i=0;i<ii;i++){                    if(ss1<ss2[i]&&pos<ss2[i]&&pos<ss1)  {flag=true; break;}                }                if(ff&&mp[9]&&flag) {puts("YES"); printf("%d92\n",tx);return 0;}            }            if(mp[4]){                bool ff=false,flag=false;                int tx=0,pos=0;                int ss1,ss2[maxn]={0},ii=0;                for(int i=0;i<len;i++){                    if(a[i]-'0'==4) ss2[ii++]=i;                }                for(int i=0;i<len;i++){                    if((a[i]-'0')%2) {tx=a[i]-'0'; pos=i; ff=true; break;}                }                for(int i=0;i<len;i++){                    if(a[i]-'0'==2) {ss1=i; break;}                }                for(int i=0;i<ii;i++){                    if(ss1<ss2[i]) {flag=true; break;}                }                if(mp[2]&&flag) {puts("YES"); printf("24\n");return 0;}                flag=false;                for(int i=0;i<len;i++){                    if(a[i]-'0'==6) {ss1=i; break;}                }                for(int i=0;i<ii;i++){                    if(ss1<ss2[i]) {flag=true; break;}                }                if(mp[6]&&flag) {puts("YES"); printf("64\n");return 0;}                flag=false;                for(int i=0;i<ii;i++){                    if(pos<ss2[i]) {flag=true; break;}                }                if(ff&&mp[4]>=2&&flag) {puts("YES"); printf("%d44\n",tx);return 0;}                flag=false;                for(int i=0;i<len;i++){                    if(a[i]-'0'==8) {ss1=i; break;}                }                for(int i=0;i<ii;i++){                    if(ss1<ss2[i]&&pos<ss2[i]&&pos<ss1) {flag=true; break;}                }                if(ff&&mp[8]&&flag) {puts("YES"); printf("%d84\n",tx);return 0;}            }            if(mp[6]){                bool ff=false,flag=false;                int tx=0,pos=0;                int ss1,ss2[maxn]={0},ii=0;                for(int i=0;i<len;i++){                    if(a[i]-'0'==6) ss2[ii++]=i;                }                for(int i=0;i<len;i++){                    if((a[i]-'0')%2) {tx=a[i]-'0'; pos=i; ff=true; break;}                }                for(int i=0;i<len;i++){                    if(a[i]-'0'==1) {ss1=i; break;}                }                for(int i=0;i<ii;i++){                    if(ss1<ss2[i]) {flag=true; break;}                }                if(mp[1]&&flag) {puts("YES"); printf("16\n");return 0;}                flag=false;                for(int i=0;i<len;i++){                    if(a[i]-'0'==5) {ss1=i; break;}                }                for(int i=0;i<ii;i++){                    if(ss1<ss2[i]) {flag=true; break;}                }                if(mp[5]&&flag) {puts("YES"); printf("56\n");return 0;}                flag=false;                for(int i=0;i<len;i++){                    if(a[i]-'0'==9) {ss1=i; break;}                }                for(int i=0;i<ii;i++){                    if(ss1<ss2[i]) {flag=true; break;}                }                if(mp[9]&&flag) {puts("YES"); printf("96\n");return 0;}                for(int i=0;i<len;i++){                    if(a[i]-'0'==3) {ss1=i; break;}                }                for(int i=0;i<ii;i++){                    if(ss1<ss2[i]&&pos<ss2[i]&&pos<ss1) {flag=true; break;}                }                if(ff&&mp[3]&&flag) {puts("YES"); printf("%d36\n",tx);return 0;}                flag=false;                for(int i=0;i<len;i++){                    if(a[i]-'0'==7) {ss1=i; break;}                }                for(int i=0;i<ii;i++){                    if(ss1<ss2[i]&&pos<ss2[i]&&pos<ss1) {flag=true; break;}                }                if(ff&&mp[7]&&flag) {puts("YES"); printf("%d76\n",tx);return 0;}            }            puts("NO");        }    }    return 0;}

感觉就像是一个暴暴力~~~

还有每次必说的——加油！