hdu1028母函數||dp

hdu1028

好像母函數的題目都挺簡單的，寫了幾個都是模板

另外還有動態規劃的方法

#include<iostream>using namespace std;int main(){int X[121],c1[121],c2[121],N,i,j,num;for(i=1;i<=120;i++)X[i]=120;while(scanf("%d",&N)!=EOF){memset(c1,0,sizeof(c1));memset(c2,0,sizeof(c2));for(i=0;i<=120;i++)c1[i]=1;for(i=2;i<=N;i++){for(j=0;j<=N;j++){for(num=0;num<=120;num++){if(num*i+j<=N)c2[num*i+j]+=c1[j];elsebreak;}}for(j=1;j<=N;j++)c1[j]=c2[j];memset(c2,0,sizeof(c2));}printf("%d\n",c1[N]);}return 0;}

改動一下

#include<iostream>using namespace std;int main(){int c1[121],c2[121],N,i,j,num;//for(i=1;i<=120;i++)X[i]=120;while(scanf("%d",&N)!=EOF){memset(c1,0,sizeof(c1));memset(c2,0,sizeof(c2));for(i=0;i<=120;i++)c1[i]=1;for(i=2;i<=N;i++){for(j=0;j<=N;j++)for(num=0;num+j<=N;num+=i)c2[num+j]+=c1[j];for(j=1;j<=N;j++)c1[j]=c2[j];memset(c2,0,sizeof(c2));}printf("%d\n",c1[N]);}return 0;}

dp如下

#include<iostream>using namespace std;int main(){int a[121],i,j,n;memset(a,0,sizeof(a));a[0]=1;for(i=1;i<=120;i++)for(j=0;i+j<=120;j++)a[i+j]+=a[j];while(scanf("%d",&n)!=EOF)printf("%d\n",a[n]);return 0;}