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
2
5 6 -1 5 4 -7
7 0 6 -1 1 -6 7 -5
Sample Output
Case 1:
14 1 4
Case 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;
}
