题目的意思就是找出未能及时处理的犯罪数，
#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; }
Codeforces Round #244 (Div. 2) A. Police Recruits,布布扣,bubuko.com
Codeforces Round #244 (Div. 2) A. Police Recruits
原文：http://www.cnblogs.com/xiongqiangcs/p/3705460.html