CodeForces 258B Little Elephant and Elections

标签（空格分隔）： 数位DP 背包DP

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算出有i个幸运数字的方案,然后枚举上限做背包DP

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
typedef long long ll;
ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
const int mod=1e9+7;
int m,f[15][15],g[15],dit[15];
ll calc(int x)
{
    ll res=1;
    for(int i=0;i<6;i++)res=res*(x-i)%mod;
    return res;
}
void MOD(int &x,int y)
{
    x+=y;
    if(x>=mod)x-=mod;
}
int dp(int step,int r,int lim)
{
    if(step==0)return r==0;
    if(!lim&&~f[step][r])return f[step][r];
    int x=lim?dit[step]:9,res=0;
    for(int i=0;i<=x;i++)
    {
        if(i==4||i==7)
        {
            if(r==0)continue;
            MOD(res,dp(step-1,r-1,lim&&i==x));
            continue;
        }
        res+=dp(step-1,r,lim&&i==x);
    }
    if(!lim)f[step][r]=res;
    return res;
}
int dfs(int step,int r,ll mul)
{
    if(step==0)return mul;
    int res=0;
    for(int i=0;i<=r;i++)if(g[i])
    {
        g[i]--;
        MOD(res,dfs(step-1,r-i,mul*(g[i]+1)%mod));
        g[i]++;
    }
    return res;
}
int main()
{
    memset(f,-1,sizeof(f));
    m=read();
    int len=0;
    while(m)dit[++len]=m%10,m/=10;
    for(int i=0;i<=len;i++)g[i]=dp(len,i,1);
    g[0]--;
    int ans=0;
    for(int i=1;i<=len;i++)MOD(ans,(ll)g[i]*dfs(6,i-1,1)%mod);
    printf("%d\n",ans);
    return 0;
}