标签:accept problem miss art ble stat str span 例子
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 164592 Accepted Submission(s): 38540
#include<iostream> using namespace std; int main() { int T;//T为输入例子的个数 cin>>T; for(int i=1;i<=T;i++) { int N;//N为每个例子中数字的个数 cin>>N; int s=0,start1=1,start2,end,maxsum=INT_MIN; //s为以目前为止的点作为结束的一类中,最大的和 ;
//start1为以目前为止的点作为结束的一类中最大和序列的起点坐标
//start2为最终确定的最大和序列中的起点坐标
//end为最终确定的最大和序列结束点的坐标
//maxsum为最终确定的最大和 for(int j=1;j<=N;j++) { int a; cin>>a; if(s<0) { s=a; start1=j; } else { s+=a; } if(s>maxsum) { maxsum=s; start2=start1; end=j; } } cout<<"Case "<<i<<":"<<endl; cout<<maxsum<<" "<<start2<<" "<<end<<endl; if(i<T) cout<<endl; } }
具体原理可参考http://alorry.blog.163.com/blog/static/6472570820123801223397/
HDU acm 1003 Max Sum || 动态规划求最大子序列和详解
标签:accept problem miss art ble stat str span 例子
原文地址:http://www.cnblogs.com/denghui666/p/7157457.html