題目描述
給定正整數N,求N分解成若干個2的次冪的方案數。
解題思路
之前寫過
將N表示成一個二進制數,對於每個
我們考慮如何構造區間
設
很快得到遞推式
於是這題效率變成
#include<cstdio>
#include<cstring>
#include<algorithm>
#define LL long long
using namespace std;
const int maxn=105,maxm=150,tt=1e9;
struct jz{
LL x[maxm];
}f[maxn][maxn],g[maxn][maxn],ans;
int n,m,a[maxn],tot;
char s[maxn];
LL b[maxn];
inline int max(int x,int y){if (x>y) return x;return y;}
jz add(jz a,jz b){
jz c;memset(c.x,0,sizeof(c.x));
c.x[0]=max(a.x[0],b.x[0]);
for (int i=1;i<=c.x[0];i++){
c.x[i]+=a.x[i]+b.x[i];
if (c.x[i]>=tt) c.x[i+1]++,c.x[i]-=tt;
}
if (c.x[c.x[0]+1]>0) c.x[0]++;
return c;
}
jz mul(jz a,jz b){
jz c;memset(c.x,0,sizeof(c.x));
c.x[0]=a.x[0]+b.x[0]-1;
for (int i=1;i<=a.x[0];i++)
for (int j=1;j<=b.x[0];j++){
c.x[i+j-1]+=a.x[i]*b.x[j];
c.x[i+j]+=c.x[i+j-1]/tt;
c.x[i+j-1]%=tt;
}
if (c.x[c.x[0]+1]>0) c.x[0]++;
return c;
}
void work(){
int j=1;
while(j<=m){
a[++n]=b[m]%2;
for (int i=j;i<=m;i++){
b[i+1]=(b[i]%2*10+b[i+1])%tt;
b[i]>>=1;
}
if (!b[j]) j++;
}
}
int main(){
freopen("exam.in","r",stdin);
freopen("exam.out","w",stdout);
scanf("%s",s+1);m=strlen(s+1);
for (int i=1;i<=m;i++) b[i]=s[i]-48;work();
for (int i=0;i<=n;i++){
for (int j=0;j<i;j++)
for (int k=0;k<=j;k++)
f[i][j]=add(f[i][j],mul(f[i-1][k],f[i-1-k][j-k]));
f[i][i].x[0]=f[i][i].x[1]=1;
}
for (int i=1;i<=n;i++)if (a[i]){
tot++;
if (tot==1) for (int j=0;j<=i;j++) g[1][j]=f[i-1][j];else
for (int j=0;j<i;j++)
for (int k=0;k<=j;k++)
g[tot][j]=add(g[tot][j],mul(g[tot-1][k],f[i-1-k][j-k]));
}
ans.x[0]=1;for (int i=0;i<=n;i++) ans=add(ans,g[tot][i]);
printf("%d",ans.x[ans.x[0]]);
for (int i=ans.x[0]-1;i;i--)
for (int j=1e8;j;j/=10) printf("%d",ans.x[i]/j),ans.x[i]%=j;
return 0;
}