Max sum HUD1003(dp)

Max Sum

Problem Description

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.

 

Input

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).

 

Output

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.

 

Sample Input

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

 

Sample Output

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

 

题目简单，基础的dp，难点在于记录子序列的开始和结束位置；

#include <stdio.h>#include <string.h>#include <stdlib.h>#include <iostream>#include <algorithm>using namespace std;struct DP{    int s, e, val;//s：起点；e：终点；val：子串的和；}dp[100005];int a[100005];int main(){    int T;    scanf("%d", &T);    int cnt=0;    while(T--){        cnt++;        int n;        scanf("%d",&n);        for(int i=1; i<=n; i++){            scanf("%d",&a[i]);            dp[i].val=a[i];    //初始化和，每个元素都是一个子串；            dp[i].e=i;        //初始化终点为i；        }        int s, e, maxx, k;        maxx=dp[1].val;        k=1;        dp[1].s=1;        dp[1].e=1;        for(int i=2; i<=n; i++)//dp[i]=max(dp[i-1]+a[i], a[i]);        {            if(dp[i-1].val+a[i] >=a[i]){                dp[i].val=dp[i-1].val+a[i];                dp[i].s=dp[i-1].s;            }            else{                dp[i].val=a[i];                dp[i].s=i;            }            if(maxx<dp[i].val){                maxx=dp[i].val;                k=i;            }        }        printf("Case %d:\n",cnt);        printf("%d %d %d\n",maxx,dp[k].s,dp[k].e);        if(T!=0) printf("\n");    }    return 0;}