Codeforces Round #244 (Div. 2) A. Police Recruits

时间:2014-05-04 11:54:32   收藏:0   阅读:274

题目的意思就是找出未能及时处理的犯罪数,

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#include <iostream>
using namespace std;

int main(){
    int n;
    cin >> n;
    int a,recruit = 0, crimes = 0;;
    for(int i = 0 ; i < n; ++ i){
        cin >> a;
        if(a > 0) recruit+=a;
        else recruit?recruit-- : crimes++;
    }
    cout<<crimes<<endl;
}
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Codeforces Round #244 (Div. 2) A. Police Recruits,布布扣,bubuko.com

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