HDOJ -- 1003 Max Sum

Max Sum

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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 

解题报告：在求一个数列的子序列的最大和时，我们需要从第一个非负数开始（temp），逐渐加上后面的数存在sum中，如果sum也为负数，则就再次从下一个非负数开始累加。每次得到的和与max比较，用max记录最大和。据说这用到了DP的思想。。。

#include<cstdio>int main(){int t,n,cnt=1;scanf("%d",&t);while(t--){int a,max;//子序列最大和 int temp;//暂存初始位置 int sum;//从temp开始到n的序列和int start,end;//记录初末位置 max=-1000;temp=1;sum=0;scanf("%d",&n);for(int i=1;i<=n;i++){scanf("%d",&a);sum+=a; if(sum>max){//如果sum大于max，更新相应的值 max=sum;start=temp;end=i;}if(sum<0){sum=0;//如果前面的数相加的和是负的，则sum清零，从下一个数开始 temp=i+1;}}printf("Case %d:\n%d %d %d\n",cnt++,max,start,end);if(t!=0)printf("\n");}return 0;}