zoj 2592 Think Positive(5-F)
2013-07-23 00:42
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题目地址:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1592
要点:1 结果就是所有数的和(如果小于0,就返回0)
2将所有的数排在圆周上(从a1到an顺时针存放),这样所谓的j,就是向顺时针扫描 不管扫描到哪里扫过的和必须都是正数。
在1和-1数目相等的时候刚好是0个j满足,然后增加一个1时可以用反证法证明恰好存在一个j,用数学归纳法可以证明: 设和为n时,j的个数为 f(n),f(n)=n
要点:1 结果就是所有数的和(如果小于0,就返回0)
2将所有的数排在圆周上(从a1到an顺时针存放),这样所谓的j,就是向顺时针扫描 不管扫描到哪里扫过的和必须都是正数。
在1和-1数目相等的时候刚好是0个j满足,然后增加一个1时可以用反证法证明恰好存在一个j,用数学归纳法可以证明: 设和为n时,j的个数为 f(n),f(n)=n
#include<iostream> using namespace std; int main() { int size; cin>>size; for(int l=0;l<size;l++) { int n; cin>>n; int sum=0; for(int i=0;i<n;i++) { int temp; cin>>temp; sum+=temp; } if(sum>=0) cout<<sum<<endl; else cout<<0<<endl; } }
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