GCJ 2015-Qualification-A Standing Ovation 难度:0

https://code.google.com/codejam/contest/6224486/dashboard#s=p0

肯定把朋友们都设置在第0位,在第i位前面必须至少有i个人鼓掌,所以答案加上i-sum[i-1]即可(sum是从第0位开始到第i-1位结束包括朋友的总人数)

#include <cstdio>

#include <cstring>

#include <algorithm>

using namespace std;

const int maxn=1e4;

int n;

int sum[maxn];

char buf[maxn];



int main(){

    freopen("data.in","r",stdin);

    freopen("data.out","w",stdout);

    int T;

    scanf("%d",&T);

    for(int ti=1;ti<=T;ti++){

        scanf("%d%s",&n,buf);

        memset(sum,0,sizeof(sum));

        int ans=0;

        sum[0]=buf[0]-'0';

        for(int i=1;i<=n;i++){

            if(sum[i-1]<i){

                ans+=i-sum[i-1];

                sum[i-1]=i;

            }

            sum[i]=sum[i-1]+buf[i]-'0';

        }

        printf("Case #%d: %d\n",ti,ans);

    }

    return 0;

}