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Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 211310 Accepted Submission(s):
49611
#include <iostream> #include <cstdio> using namespace std; int main() { int t; int n; int a[100005];//存储序列 int sum[100005];//存储以每个数为结尾的子序列和 int s[100005];//存储开始位置 int ans;//结束位置 scanf("%d",&t); for(int i=1;i<=t;i++){ scanf("%d",&n); for(int i1=0;i1<n;i1++){ scanf("%d",&a[i1]); } ans=0; sum[0]=a[0]; s[0]=0; for(int j=1;j<n;j++){ if(sum[j-1]>=0){ sum[j]=sum[j-1]+a[j]; s[j]=s[j-1]; }else{ sum[j]=a[j]; s[j]=j; } if(sum[ans]<sum[j]) ans=j; } if(i<t){ printf("Case %d:\n%d %d %d\n",i,sum[ans],s[ans]+1,ans+1); printf("\n"); }else{ printf("Case %d:\n%d %d %d\n",i,sum[ans],s[ans]+1,ans+1); } } return 0; }
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原文地址:http://www.cnblogs.com/TWS-YIFEI/p/5590532.html