HDU--Max sum（DP）

来源：互联网发布：沉默的螺旋理论知乎编辑：程序博客网时间：2024/05/21 09:59

Max Sum

Time Limit: 2000ms Memory limit: 32768K 有疑问？点这里^_^

题目描述

Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the max sum in this sequence is 6 + (-1) + 5 + 4 = 14.

输入

The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line starts with a number N(1<=N<=100000), then N integers followed(all the integers are between -1000 and 1000).

输出

For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line contains three integers, the Max Sum in the sequence, the start position of the sub-sequence, the end position of the sub-sequence. If there are more than one result, output the first one. Output a blank line between two cases.

示例输入

25 6 -1 5 4 -77 0 6 -1 1 -6 7 -5

示例输出

Case 1:14 1 4Case 2:7 1 6

提示

hdoj1003 注意：本题后台测试数据量比较大，请使用C语言scanf输入。避免超时！！！！

又是一道很经典的dp，题意是让求最大连续字段和不难写出状态转移方程 dp[i]=max(dp[i-1]+i,i);

#include <cstdio>int a[100001];int main(){ int t,n,i,j,k; scanf("%d",&t); for(i=1;i<=t;i++) {  scanf("%d",&n);  for(j=0;j<n;j++)  scanf("%d",&a[j]);  int sum=0,max=-99999,k=0,pos1,pos2;  for(j=0;j<n;j++)  {   sum+=a[j];   if(sum>max)   {    max=sum;pos1=k;pos2=j;   }   if(sum<0)   {    k=j+1;sum=0;   }  }  printf("Case %d:\n",i);  if(i!=t)  printf("%d %d %d\n\n",max,pos1+1,pos2+1);  else  printf("%d %d %d\n",max,pos1+1,pos2+1); } return 0;}

2 0