Hdoj 1003.Max Sum 題解
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
思路
最大連續子序列和問題,狀態轉移方程式:
\(f[i] = max(f[i-1]+a[i],a[i])\)
可以得出代碼如下
代碼
#include<bits/stdc++.h> using namespace std; const int INF = 1<<30; int a[100001]; int main() { int n; cin >> n; for(int q=1;q<=n;q++) { int len; cin >> len; int maxsum = -INF; int currentsum = 0; int l = 0,r = 0; int tmp = 1; for(int i=1;i<=len;i++) { cin >> a[i]; if(currentsum >= 0) currentsum += a[i]; else { currentsum = a[i]; tmp = i; } if(currentsum > maxsum) { maxsum = currentsum; l = tmp; r = i; } } cout << "Case " << q << ":\n"; cout << maxsum << " " << l << " " << r << endl; if(q!=n) cout << endl; } return 0; }
Hdoj 1003.Max Sum 題解