Hdu 1003 最大子段和

http://acm.hdu.edu.cn/showproblem.php?pid=1003

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 246205    Accepted Submission(s): 58135

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

题意：给出一组长度在100000以内的数字（-1000~1000），求这些数字里连续的子段和最大是多少，并输出子段的头和尾。

分析：可以用两层循环来计算所有子序列的和，再从这些和里找出最大的，但是这样会超时。

从前往后扫，当和大于max时更新尾标记，当和小于0时将头标记移到现在扫到的下一位，将sum重置为0，继续计算和。

#include<stdio.h>int main(){int n,i,j,T,sum,max,t,head,tail;int a[100010];scanf("%d",&T);for(t=1;t<=T;t++){sum=0;head=tail=j=0;max=-1010;scanf("%d",&n);for(i=0;i<n;i++)scanf("%d",&a[i]);for(i=0;i<n;i++){sum+=a[i];if(sum>max){head=j+1;//更新头部标记tail=i+1;//更新尾部标记max=sum;}if(sum<0){ sum=0;j=i+1;}}printf("Case %d:\n",t);printf("%d %d %d\n",max,head,tail);if(t<T)putchar('\n');}return 0; }