hdu 1003 (动态规划入门)Max Sum

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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
25 6 -1 5 4 -77 0 6 -1 1 -6 7 -5
 

Sample Output
Case 1:14 1 4Case 2:7 1 6
#include <iostream>#include <cstdio>using namespace std;const int maxn=100000+10;int main(){int n,T,a[maxn],max,begin,end,k,i,t,j;scanf("%d",&n);for (j=1;j<=n;j++){t=0;max=-9999;k=1;//max的初始值很重要,因为子串中可能有负数scanf("%d",&T);for (i=1;i<=T;i++){scanf("%d",a+i);t+=a[i];if (t>max)//找到更大的一个子串,更新最大值和起点,终点{max=t;begin=k;end=i;}if (t<0)//可能还有比之前大的子串{t=0;k=i+1;//不能把i+1赋值给begin,因为不一定下个子串和最大}}printf("Case %d:\n",j);if (j!=n){printf("%d %d %d\n\n",max,begin,end);}elseprintf("%d %d %d\n",max,begin,end);}return 0;}

难点开始了,fighting!!!

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