hduoj1003(经典dp入门)

Max Sum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 207382 Accepted Submission(s): 48503

Problem Description

Given a sequence a[1],a[2],a[3]……a
, 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

dp[i]为到i为止最大的和，dp[i]=max(dp[i-1]+a[i],a[i])

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
typedef int  ll;
#define MIN -9999
ll max(ll a,int b){
return a>b?a:b;
}
ll dp[100005];
ll ans,ans1;
int a[100005];
int main(){
int t,n,i,j,e,f;
//freopen("input.txt","r",stdin);
scanf("%d",&t);
for(i=1;i<=t;i++){
//ans=MIN;
//memset(dp,MIN,sizeof(dp));
scanf("%d",&n);
for(j=1;j<=n;j++)
scanf("%d",&a[j]);
e=1;
ans=dp[1]=a[1];
for(j=2;j<=n;j++){
dp[j]=max(dp[j-1]+a[j],a[j]);
if(ans<dp[j]){
ans=dp[j];
e=j;
}
}
ans1=0;
f=e;
for(j=e;j>0;j--){
ans1+=a[j];
//printf("%d ",j);
if(ans1==ans){
//printf("ans\n");
f=j;
}
}

printf("Case %d:\n%d %d %d\n",i,ans,f,e);
if(i!=t)printf("\n");

}
return 0;
}