Max Sum
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 81735 Accepted Submission(s): 18797
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
解题心得:
1、这个题要求的是一个在序列中的一串连续的子序列,要求子序列的和最大,并且要求记录子序列的起始位置和终止位置。
2、可以将前面的数加在一起看作一个数,当前面的数为负数的时候则舍去,将当前这个数作为新的起点。用一个变量Max记录一下最大的值,当最大值被替换的时候记录一下终点和起点。理解了还是很容易的。
#include<bits/stdc++.h> using namespace std; const int maxn = 1e5+100; int num[maxn],dp[maxn]; int Start,End,now_start,Max; int main() { int t; int T; scanf("%d",&t); T = t; while(t--) { memset(dp,0,sizeof(dp)); num[0] = -1; now_start = 1; Max = -1000000; int n; scanf("%d",&n); for(int i=1;i<=n;i++) scanf("%d",&num[i]); for(int i=1;i<=n;i++) { if(num[i] + dp[i-1] >= num[i])//前面的数不是负数 dp[i] = num[i] + dp[i-1]; else { dp[i] = num[i];//前面一个数是负数,将现在这个数作为新的起点 now_start = i;//当前起点,当出现最大值的用得到 } if(dp[i] >= Max)//出现最大值,记录最大值,终点和起点 { Start = now_start; End = i; Max = dp[i]; } } printf("Case %d: ",T-t); if(t != 0) printf("%d %d %d ",Max,Start,End); else printf("%d %d %d ",Max,Start,End); } return 0; }