HDU1003`Max Sum(DP)

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 244914    Accepted Submission(s): 57821

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

给出n个数，求连续子串和的最大值，并记录起始点和终止点

直接遍历一遍，记录前缀和并不断刷新最大值和右端点，如果前缀和<0，将记录值变为0，左端点变为此时的下一点

#include <iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#define inf 0x3f3f3f3f#define mem(a,b) memset(a,b,sizeof(a))#define N 100005using namespace std;int a[N];int main(){    int t,q=1;    scanf("%d",&t);    while(t--)    {        int n;       scanf("%d",&n);       for(int i=1;i<=n;i++)        scanf("%d",&a[i]);       int cnt=0,maxn=-1000,x1=1,x,y;       for(int i=1;i<=n;i++)       {           cnt+=a[i];           if(cnt>maxn)            maxn=cnt,x=x1,y=i;           if(cnt<0)            cnt=0,x1=i+1;       }       printf("Case %d:\n%d %d %d\n",q++,maxn,x,y);       if(t)        puts("");    }}