UOJ275 组合数问题

给定n,m和k,求有多少对(i , j)满
足0 ≤ i ≤ n, 0 ≤ j ≤ min(i ,m)且C(︀i，j)︀是k的倍数.
n,m ≤ 1018, k ≤ 100,且k是质数.

把i和j都看成k进制数，事实上这个问题就是问有多少
对j ≤ i满足j有一位比i大.

转化成数位DP

对每一位进行转移

bool判断是否当前位n<=a[i],m<=b[i]

By：大奕哥

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int mod=1e9+7;
ll inv=500000004ll;
int T;
ll f[70][2][2],a[70],b[70],k;

ll calc(ll x,ll y)
{
    if(x<0||y<0)return 0;
    if(x<y)return 1ll*((x+2)%mod*(x+1)%mod)%mod*inv%mod;
    return (1ll*(y+2)%mod*((y+1)%mod)%mod*inv%mod+1ll*(x-y)%mod*((y+1)%mod)%mod)%mod;
}
int main()
{
    scanf("%d%lld",&T,&k);
    while(T--)
    {
        int n=0,m=0;long long x,y;
        scanf("%lld",&x);ll t=x;
        while(t)
        {
            a[++n]=t%k;t/=k;
        }
        scanf("%lld",&y);y=min(y,x);t=y;
        while(t)
        {
            b[++m]=t%k;t/=k;
        }
        long long ans=calc(x,y);
        f[0][1][1]=1;
        for(int i=1;i<=n;++i)
        {
            f[i][1][1]=(calc(a[i],b[i])*f[i-1][1][1]%mod+calc(a[i],b[i]-1)*f[i-1][1][0]%mod+calc(a[i]-1,b[i])*f[i-1][0][1]%mod+calc(a[i]-1,b[i]-1)*f[i-1][0][0]%mod)%mod;
            f[i][0][1]=(calc(k-1,b[i])*(f[i-1][1][1]+f[i-1][0][1])%mod+calc(k-1,b[i]-1)*(f[i-1][1][0]+f[i-1][0][0])%mod-f[i][1][1]+mod)%mod;
            f[i][1][0]=(calc(a[i],k-1)*(f[i-1][1][1]+f[i-1][1][0])%mod+calc(a[i]-1,k-1)*(f[i-1][0][1]+f[i-1][0][0])%mod-f[i][1][1]+mod)%mod;
            f[i][0][0]=(((calc(k-1,k-1)*(f[i-1][1][1]+f[i-1][1][0]+f[i-1][0][1]+f[i-1][0][0])%mod-f[i][1][1]+mod)%mod-f[i][0][1]+mod)%mod-f[i][1][0]+mod)%mod;
        }
        printf("%lld\n",(ans-f[n][1][1]+mod)%mod);
        while(n)a[n--]=0;
        while(m)b[m--]=0;
    }
}