poj 1276 Cash Machine
這是一個簡單的dp題,用給定數量和麵值的錢去換儘量接近所給的錢的大小。
第一次是用三重迴圈來做AC了感覺效率不是很高,
10448082 hellobabygogo 1276 Accepted 640K 516MS C++ 595B 2012-07-14 17:48:59程式碼:
#include<iostream>
using namespace std;
int cash,N,d[11],n[11],dp[100000];
int main()
{
int i,j,k;
while(cin>>cash>>N)
{
for(i=0;i<N;i++)
cin>>n[i]>>d[i];
memset(dp,0,sizeof(dp));
dp[0]=1;
for(i=0;i<N;i++)
{
for(j=cash;j>=0;j--)
if(dp[j])
{
for(k=1;k<=n[i];k++)
{
if(j+k*d[i]>cash) break;
dp[j+k*d[i]]=1;
}
}
}
for(i=cash;i>=0;i--)
if(dp[i]) {cout<<i<<endl;break;}
}
return 0;
}
做了優化,把三重迴圈改成兩重迴圈。
10448802
hellobabygogo 1276 Accepted 948K 0MS C++ 838B 2012-07-14 20:06:23
程式碼:
#include<iostream>
using namespace std;
int cash,N,d[11],n[11],dp[100010],num[100010];
int main()
{
int i,j,k;
while(cin>>cash>>N)
{
for(i=0;i<N;i++)
cin>>n[i]>>d[i];
memset(dp,0,sizeof(dp));
dp[0]=1;
for(i=0;i<N;i++)
{
memset(num,0,sizeof(num));
for(j=d[i];j<=cash;j++)
if(!dp[j]&&dp[j-d[i]]&&num[j-d[i]]<n[i])
{dp[j]=1;num[j]=num[j-d[i]]+1;}
}
for(i=cash;i>=0;i--)
if(dp[i]) {cout<<i<<endl;break;}
}
return 0;
}
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