Max Sum
页面数据来自(this page from)： http://acm.hdu.edu.cn/showproblem.php?pid=1003
- Time Limit: 2000/1000 MS (Java/Others)
- Memory Limit: 65536/32768 K (Java/Others)
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
Author
Ignatius.L
Statistic | Submit | Discuss | Note
Source Code
#include <iostream>
using namespace std;
int max(int a,int b){ return a>b?a:b; }
int main(){
int T,N;
cin >> T;
for(int i=0;i<T;i++){
int f[100001],start=0,end=0,big=-1000;
cin >> N;
cin >> f[0];
big = max(f[0],big);
for(int j=1;j<N;j++){
cin >> f[j];
//求和
int sum = f[j-1] + f[j] ;
//如果 sum 大于 f[j] 的本身，且大于0，那么视为增益效果
if(sum >= f[j] && sum >= 0){
f[j] = sum;
}
big = max(f[j],big);
}
//寻找开始和结束的下标
// f[j] 小于0则视为增益区间结束
// f[j] 大于0则为增益区间
for(int j=0;j<N;j++){
if(f[j]==big){
end = j;
break;
}
if(f[j]<0) start=j+1;
}
cout << "Case " << i+1 << ':' << endl;
cout << big << ' ' << start+1 << ' ' << end+1 <<endl;
if(i<T-1)cout <<endl;
}
return 0;
}
自己摸索的算法，不太会 dp 和贪心，大佬们勿喷。
