HDU-1003 Max Sum (dp)
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题目:
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 4Case 2:
7 1 6
思路:
经典入门dp,难度小,主要因为自己太粗心WA了4次,记录一下长记性。
WA的地方是记录子序列末位置last时没有初始化为1,导致如果最大子序列就是第一个元素的时候last为0。
Code:
#include<bits/stdc++.h>using namespace std;#define MAXN 100000+10#define M(a, b) memset(a, b, sizeof(a))int a[MAXN], dp[MAXN], first, last, maxsum;void Do(){ M(dp, 0); M(a, 0); int n; scanf("%d", &n); for (int i = 1; i <= n; ++i) scanf("%d", &a[i]); dp[1] = a[1]; for (int i = 2; i <= n; ++i) dp[i] = max(a[i]+dp[i-1], a[i]); maxsum = dp[1]; last = 1; for (int i = 2; i <= n; ++i) if (dp[i] > maxsum){ maxsum = dp[i]; last = i; } int temp = maxsum; for (int i = last; i >= 1; --i){ temp -= a[i]; if (!temp) first = i; }}int main(){ int T; scanf("%d", &T); for (int i = 1; i <= T; ++i){ Do(); printf("Case %d:\n%d %d %d\n", i, maxsum, first, last); if (i != T) printf("\n"); } return 0;}
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