HDU 1003 区间最大和问题（动态规划）

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Max Sum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 183319 Accepted Submission(s): 42766

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动态规划问题，给你一个数组，让你求从一个从i到j和最大的这么一个范围，策略是：每输入第i个数要去判断要不要把这个数加入我前面所组成的数组x[i-1]里，如果加入这个数后得到和不小于这个数，那就把这个数加入前面的数列，否则重新开始构建最大和的数列。用两个数组x[i]存max(x[i]+a,a),x1[i]存处理每一个数时开始的位置例子：输入 5   6 -1 5 4 7时 6-154-7x6510147x111111循环一遍找出最大的和为14，开始的位置为1，结束位置为4（第4个数）例子输入 8  0 6 -7 1 6 1 -5 9 （自己造的例子） 06-7161-59x06-1178312x111144444（输入第4个数的时候（-1+1=0<1）所以重新从第四个数开始）循环一遍找出最大的和为12，开始的位置为4，结束位置为8（第8个数）Accepted代码#include<iostream>#include<cstdio>#include<cstring>using namespace std;const int INF=0x3f3f3f3f;int main(){    int T;    int x[100005];    int x1[100005];    int cas=1;    scanf("%d",&T);    while(T--)    {        memset(x,0,sizeof(x));        memset(x1,0,sizeof(x1));        //memset(x2,0,sizeof(x2));        int a,b=-INF,n;        scanf("%d",&n);        x[0]=-INF;        x1[0]=1;        for(int i=1;i<=n;i++)        {            scanf("%d",&a);            if(x[i-1]+a>=a)            {                x[i]=x[i-1]+a;                x1[i]=x1[i-1];            }            //x[i]=max(x[i-1]+a,a);            else            {                x[i]=a;                x1[i]=i;            }        }        int s,e;        for(int i=1;i<=n;i++)        {            if(x[i]>b)            {                b=x[i];                s=x1[i];                e=i;            }        }        printf("Case %d:\n",cas++);        printf("%d %d %d\n",b,s,e);        if(T!=0)printf("\n");    }}

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