統計硬幣
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3846 Accepted Submission(s): 2695
接下來的T行,每行有兩個數n,m,n和m的含義同上。
每組輸出佔一行。
AC CODE:
#1:非母函數法
#include<stdio.h>
//int c1[1000],c2[1000];
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int m,n;
scanf("%d%d",&n,&m);
int i,j,k,sum,temp=0;
for(i=0;i<=n;i++)
for(j=0;j<=n-i;j++)
for(k=0;k<=n-i-j;k++)
{
if(i+k+j!=n) continue;
sum=i+2*j+5*k;
if(sum==m)
temp++;
}
printf("%d\n",temp);
}
return 0;
}
#2:母函數法
#include<stdio.h>
#include<string.h>
int c1[100][100],c2[100][100];
int main()
{
int i,j,k,l,n,m,t;
scanf("%d",&t);
while(t--)
{
int sum=0;
scanf("%d%d",&n,&m);
memset(c1,0,sizeof(c1));
memset(c2,0,sizeof(c2));
for(i=0;i<=n;i++)
{
c1[i][i]=1;
}
for(i=2;i<=5;i+=3)
{
for(j=0;j<=m;j++)
for(k=0;k*i+j<=m&&k<=n;k++)
{
for(l=0;l+k<=n;l++)
c2[k*i+j][k+l]+=c1[j][l];
}
for(j=0;j<=m;j++)
{
for(k=0;k<=n;k++)
{
c1[j][k]=c2[j][k];
c2[j][k]=0;
}
}
}
printf("%d\n",c1[m][n]);
}
return 0;
}