HDU-1003-Max Sum

A - Max Sum
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
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Status

Practice

HDU 1003
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
2
5 6 -1 5 4 -7
7 0 6 -1 1 -6 7 -5

Sample Output
Case 1:
14 1 4

Case 2:
7 1 6

求最大子串和，和NYOJ-44题一样，只是这题还要输出区间
NYOJ-44http://blog.csdn.net/qq_32680617/article/details/50759863

思想都在注释里，不过这次代码写的不够简练，想的有点多。
注意空行输出的条件

#include<stdio.h>#include<string.h>#include<string>#include<stack>#include<queue>#include<math.h>#include<limits.h>#include<iostream>#include<algorithm>using namespace std;int this_i,this_j;//当前区间int max_i,max_j;//最终区间int this_sum;//当前最大值int max_sum;//最终最大值int temp;//临时接收数据int T,N;//俩没用的数据int main(){    cin>>T;    int num=1;//计数的    while(T--)    {        cin>>N;        cin>>this_sum;//第一个数据作为初始值        max_sum=this_sum;        this_i=1,this_j=1,max_i=1,max_j=1;//初始化各值        for(int i=2; i<=N; i++)        {            cin>>temp;//接收到一个数据            if((this_sum+temp)<temp)            {                //如果当前最大值加上当前数据后小于当前数据                this_sum=temp;//当前数据赋值给当前最大值                this_i=i;//更改当前区间                //this_j=i;            }            else            {                //如果当前最大值加上当前数据后大于等于当前数据                this_sum+=temp;//当前最大值累加               // this_j++;//当前区间终点+1            }            if(this_sum>max_sum)//如果当前最大值大于最终最大值            {                max_sum=this_sum;//更新最终最大值并更新最终区间                max_i=this_i;                max_j=i;            }        }        printf("Case %d:\n",num++);        printf("%d %d %d\n",max_sum,max_i,max_j);        if(T!=0)            printf("\n");    }    return 0;}

动态规划的入门题