练习三 Problem A
原题：
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
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
题目大意：求最大的连续字段和，求出并输出该子段和的起始位置。
解题思路：输入一串整数，记一个最大和summax,使其不断与前i个数的和比较若sum[i]>summax并记录开始与结束的位置,则使summax=sum[i];若sum[i]<0,则sum[i]=0并记录此时的位置为起始位置;此后不断循环最终得到所求结果。注意起始位置的记录。
感想：经典dp题，新手必做，刚开始的时候做还是有点困难。
代码：
#include
#include
using namespace std;
int main()
{
int n,t,i,c;
int a[100002];
scanf("%d",&t);
for(c=1;c<=t;c++)
{
int k=1,st=0,en=0,summax=-1000,sum=0;
scanf("%d",&n);
for(i=0;isummax)
{
summax=sum;
st=k;
en=i+1;
}
if(sum<0)
{
sum=0;
k=i+2;
}
}
printf("Case %d:\n%d %d %d\n",c,summax,st,en);
if(c!=t) cout<