Cut Ribbon
2013-08-09 08:41
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Cut Ribbon
Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d
& %I64u
SubmitStatusPracticeCodeForces
189A
Description
Polycarpus has a ribbon, its length is n. He wants to cut the ribbon in a way that fulfils the following two conditions:
After the cutting each ribbon piece should have length
a, b or
c.
After the cutting the number of ribbon pieces should be maximum.
Help Polycarpus and find the number of ribbon pieces after the required cutting.
Input
The first line contains four space-separated integers n,
a, b and
c (1 ≤ n, a, b, c ≤ 4000) — the length of the original ribbon and the acceptable lengths of the ribbon pieces after the cutting, correspondingly. The numbers
a, b and
c can coincide.
Output
Print a single number — the maximum possible number of ribbon pieces. It is guaranteed that at least one correct ribbon cutting exists.
Sample Input
Input
Output
Input
Output
Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d
& %I64u
SubmitStatusPracticeCodeForces
189A
Description
Polycarpus has a ribbon, its length is n. He wants to cut the ribbon in a way that fulfils the following two conditions:
After the cutting each ribbon piece should have length
a, b or
c.
After the cutting the number of ribbon pieces should be maximum.
Help Polycarpus and find the number of ribbon pieces after the required cutting.
Input
The first line contains four space-separated integers n,
a, b and
c (1 ≤ n, a, b, c ≤ 4000) — the length of the original ribbon and the acceptable lengths of the ribbon pieces after the cutting, correspondingly. The numbers
a, b and
c can coincide.
Output
Print a single number — the maximum possible number of ribbon pieces. It is guaranteed that at least one correct ribbon cutting exists.
Sample Input
Input
5 5 3 2
Output
2
Input
7 5 5 2
Output
2开始的时候初始化错误,错了好几次!!!
#include<iostream> using namespace std; int max(int a,int b) { return a>b?a:b; } int dp[4010]; int main() { int n,a[3]; int i,j; while(cin>>n>>a[0]>>a[1]>>a[2]) { for(i=0;i<=n;i++) { dp[i]=-4010; } dp[0]=0; for(i=0;i<3;i++) { for(j=a[i];j<=n;j++) { dp[j]=max(dp[j],dp[j-a[i]]+1); } } cout<<dp <<endl; } }