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+;
 ll inv=500000004ll;
 int T;
 ll f[][][],a[],b[],k;

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