简单DP HDOJ 1003 MAX SUM

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 223582    Accepted Submission(s): 52609

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.

顺便附下链接：http://acm.hdu.edu.cn/showproblem.php?pid=1003

这是一个DP题，首先就要去寻找子问题及其相应的策略，当我们进行一次遍历时发现，依次加和去寻找最大值时，我们需要去寻找在何种情况下开始去第一个需要加的，何时截止；经此，我们发现此问题在每次加和时的SUM取值必须大于0，否则必须从这个值开始新一轮的加和，至于截止只需进行完一次遍历后，SUM最大；

废话不多说，直接粘代码。

#include <stdio.h>int a[100010];int main(){int T,N,i,cas=1;int start,end,temp,sum,max=0;scanf("%d",&T);while(T--){scanf("%d",&N);start=end=temp=1;sum=0;max=-1001;for(i=1;i<=N;i++){scanf("%d",&a[i]);sum+=a[i];if(sum>max){max=sum;start=temp;end=i;}if(sum<0){sum=0;temp=i+1;}}printf("Case %d:\n",cas++);printf("%d %d %d\n",max,start,end);if(T>0)printf("\n");}return 0;}