Max Sum
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 141547 Accepted Submission(s): 32929
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
Author
Ignatius.L
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此题与HDU 1231一样，不过简单一些，记录前标要容易。
#include <iostream>
using namespace std;
#define M 100100
int vis[M],dp[M];
int main(int i,int j,int k)
{
int n,last,t,first,temp;
cin>>t;
for(k=1;k<=t;k++)
{
cin>>n;
memset(dp,0,sizeof(dp));
for(i=1;i<=n;i++) {cin>>vis[i];dp[i]=vis[i];}
int sum=0,maxSum=vis[1];first=last=temp=1;
for(i = 1; i <= n; i++)
{
sum += vis[i]; //i是从1开始的，所以先加再判断。
if(sum > maxSum){
maxSum = sum;
first = temp;
last = i;
}
if(sum < 0){ //这里用sum表示vis[first]->vis[i]的和，所以要清零。
sum = 0;
temp = i+1;
}
}
printf("Case %d:\n",k);
printf("%d %d %d\n",maxSum,first,last);
if(k < t)
printf("\n"); 好吧，这里就让我格式错误了几次。。。明明Case 1:和Case 2:数据中间有空行，但是完成一次没空行。。。
}
return 0;
}
HDU 1003 Max Sum (动规),布布扣,bubuko.com
HDU 1003 Max Sum (动规)
原文：http://blog.csdn.net/qq2256420822/article/details/37989659
