入门硬币问题

第一题：给出n个硬币，不同值，每一个价值的硬币有无数个，求最多硬币数刚好凑到价值为S的硬币数，，，？最少又是多少？

方法：1、用递归方法，两种情况得分别写，注意把每一个有没有计算过的值给记录。这种方法容易搞混。dp1(),dp2();

2、用二维数组迭代：注意顺序dp()

#include<iostream>
#include<cstdio>
#include<cstring>
#include<ctype.h>
#include<vector>
#include<algorithm>
#include<cmath>
#include<queue>
#include<set>
#include<string>
//#include<map>
#include<stack>
using namespace std;

#define N 100
#define Inf 100000000

int s;
int c
;
int d[N*N];

int n;

int dp1(int i)
{
//cout<<i<<endl;
int &ans=d[i];
if (ans!=-1)
return ans;

ans=-1<<30;
for (int j=1;j<=n;j++)
{
if (i>=c[j])
ans=max(ans,dp1(i-c[j])+1);
}
return ans;
}

int dp2(int i)
{
int &ans=d[i];
if (ans!=-1)
return ans;

ans=1<<30;
for (int j=1;j<=n;j++)
{
if (i>=c[j])
{
ans=min(ans,dp2(i-c[j])+1);
}
}
return ans;
}

void dp()
{
int small[N*N],big[N*N];
small[0]=0;
big[0]=0;
for (int i=1;i<=s;i++)
{
small[i]=Inf;
big[i]=-Inf;

}
int path_small[N*N];
int path_big[N*N];
for (int i=1;i<=s;i++)
{
for (int j=1;j<=n;j++)
{
if (i>=c[j])
{
if (small[i]>small[i-c[j]]+1)
{
small[i]=small[i-c[j]]+1;
path_small[i]=j;
}
if (big[i]<big[i-c[j]]+1)
{
big[i]=big[i-c[j]]+1;
path_big[i]=j;
}
}
}
}
cout<<"The big one is :   "<<big[s]<<endl;
int cur=s;
while (cur)  //打印用的是哪些硬币
{
cout<<c[path_big[cur]]<<' ';
cur-=c[path_big[cur]];
}
cout<<endl;
cout<<"The small one is:    "<<small[s]<<endl;
cur=s;
while (cur)
{
cout<<c[path_small[cur]]<<' ';
cur-=c[path_small[cur]];
}
cout<<endl;

}

void print(int k)
{

for (int i=1;i<=n;i++)
{
if (k>=c[i]&&d[k]==d[k-c[i]]+1)
{
cout<<c[i]<<' ';
print(k-c[i]);
break;
}
}

}

int main()
{
freopen("fuck.txt","r",stdin);
int i,j,k;

int num=0;
while (cin>>n>>s,n)
{
num++;

for (i=1;i<=n;i++)
cin>>c[i];
if (0)
dp();
else
{
memset(d,-1,sizeof(d));
d[0]=0;
cout<<"#Case "<<num<<" is :"<<endl;
cout<<"The big one is:  "<<dp1(s)<<endl;
print(s); //打印
cout<<endl;
memset(d,-1,sizeof(d));
d[0]=0;
cout<<"The small one is:   "<<dp2(s)<<endl;
print(s);
cout<<endl;
cout<<endl;
//打印也是从0开始找到第一个，
}
}

return 0;
}