Max Sum
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
#include <bits/stdc++.h> using namespace std; const int MAXN = 100000 + 10; int T, n; int arr[MAXN], dp[MAXN]; int S, E; int main() { scanf("%d", &T); for(int t = 0; t < T; ++t) { S = E = 0; cin >> n; for(int i = 0; i != n; ++i) cin >> arr[i]; dp[0] = arr[0]; for(int i = 1; i != n; ++i) if(dp[i-1] >= 0) dp[i] = dp[i-1] + arr[i]; else dp[i] = arr[i]; int Max = dp[0]; for(int i = 1; i != n; ++i) if(dp[i] >= Max) { Max = dp[i]; E = i; } int sum = 0; for(int i = E; i >= 0; --i) { sum += arr[i]; if(sum == Max) S = i; } cout << "Case " << t+1 << ":" << endl; cout << Max << " " << S+1 << " " << E+1 << endl; if(t < T-1) puts(""); } return 0; }
#include <bits/stdc++.h> using namespace std; int main() { int T,n; int Max, S, E, sum, a; cin >> T; for(int t = 1; t <= T; ++t) { cin >> n; S = E = sum = 0; Max = -100000; int k = 0; for(int i = 0; i != n; ++i) { cin >> a; sum += a; if(sum > Max) { Max = sum; S = k; E = i; } if(sum < 0) { sum = 0; k = i + 1; } } cout << "Case " << t << ":" << endl; cout << Max << " " << S+1 << " " << E+1 << endl; if(t < T) puts(""); } return 0; }