HDU 1003(动态规划-最大连续区间和)

问题描述:

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

题目题意:给一个数组，求最大连续区间和！

代码如下:

#include<iostream>#include<cstdio>#include<cstring>using namespace std;const int maxn=1e5+1000;int k,sum,ans,s,e,anss,anse,a[maxn];int main(){    int t,icase=0;    scanf("%d",&t);    while (t--) {        scanf("%d",&k);        for (int i=1;i<=k;i++)            scanf("%d",&a[i]);        sum=ans=a[1];        s=e=anss=anse=1;        for (int i=2;i<=k;i++) {            if (sum>=0) {                sum+=a[i];                e=i;            }            if (sum<0) {//如果前缀和小于0,那么另起起点肯定更优                sum=a[i];                s=e=i;            }            if (sum>ans) {                ans=sum;                anss=s;                anse=e;            }        }        printf("Case %d:\n",++icase);        printf("%d %d %d\n",ans,anss,anse);        if (t) printf("\n");    }    return 0;}