Codeforces Round #244 (Div. 2) A. Police Recruits
时间:2014-05-04 11:53:55
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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; }
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原文:http://www.cnblogs.com/xiongqiangcs/p/3705460.html
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