POJ- -3628 Bookshelf 2

Title Link: http://poj.org/problem?id=3628

Test instructions

Problem Solving Ideas:

1 determine the maximum possible height sum, which is the height of all the cow added up

2 According to the dynamic programming method, the possible solution between the sum of 1 to the maximum height is solved

3 found the lowest height than B (Bookshelf height), which may be the same as B.

Sample Input 5  - 3 1 3 5 6 Sample Output 1

AC Code:

1#include <stdio.h>2#include <stdlib.h>3#include <string.h>4#include <queue>5#include <algorithm>6#include <cmath>7#include <iostream>8 9 using namespacestd;TentypedefLong LongLL; One  A #defineINF 0x3f3f3f3f - #defineN 1200000 - #defineMAXN 100000000 the #defineMoD 1000000007 -  - intDp[n],a[n]; -  + intMain () - { +     intn,x,sum,i,j; A  at      while(SCANF ("%d%d", &n,&x)! =EOF) -     { -Memset (DP,0,sizeof(DP)); -Memset (A,0,sizeof(a)); -  -sum=0; in          for(i=1; i<=n;i++) -         { toscanf"%d", &a[i]); +sum+=A[i]; -         } the  *          for(i=1; i<=n;i++) $              for(j=sum;j>=a[i];j--)Panax NotoginsengDp[j]=max (dp[j],dp[j-a[i]]+a[i]); -  the          for(i=1; i<=sum;i++) +         { A             if(dp[i]>=x) the             { +printf"%d\n", dp[i]-x); -                  Break; $             } $         } -     } -     return 0; the}

POJ- -3628 Bookshelf 2