題目分析：

我記得很久以前有人跟我說NOIP2016的題目出了加強版在清華集訓中，但這似乎是一道無關的題目？

由於$k$為素數，那麽$lucas$定理就可以搬上臺面了。

註意到$\binom{i}{j} \equiv 0 {\mod k}$當且僅當將$i$和$j$用$k$進制表示的時候，有一位上的$i<j$。

位數上的計算用數位DP就沒錯了。

代碼：

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 
 4 const int mod = 1000000007;
 5 
 6 int t,k;
 7 long long n,m;
 8
 
 
 9 int bn[90],n1,n2,bm[90],nw[90];
10 
11 int f[70][100][2]; //0 0~k-1 1 0~self
12 int sum[70][2],pw[70],dd[70],oo[70],fw[70],yw[70];
13 
14 void init(){
15     if(n < m) m = n;
16     memset(bn,0,sizeof(bn)); memset(bm,0,sizeof(bm)); n1 = 0,n2 = 0;
17     long long p1 = n,p2 = m;
18     while(p1){bn[++n1] = p1 % k; p1 /= k;}
 
19     while(p2){bm[++n2] = p2 % k; p2 /= k;}
20     dd[0] = 1;oo[0] = 1;
21     for(int i=1;i<=n1;i++) dd[i] = (dd[i-1] + 1ll*bm[i]*pw[i-1]%mod)%mod;
22     for(int i=1;i<=n1;i++) oo[i] = (oo[i-1] + 1ll*bn[i]*pw[i-1]%mod)%mod;
23 }
24 
25 pair<int,int> dfs3(int now){
26     if(now == 0){return
 
 make_pair(0,0);}
27     int ans1 = 0,ans2 = 0;
28     pair<int,int> pt = dfs3(now-1);
29     int cutp = min(bn[now]+1,bm[now]);
30     for(int i=0;i<bn[now];i++){
31     int cuep = min(i+1,bm[now]);
32     ans1 += (1ll*cuep*sum[now-1][0])%mod; ans1 %= mod;
33     ans1 += (1ll*(bm[now]-cuep)*pw[now-1])%mod*pw[now-1]%mod;ans1%=mod;
34     if(bm[now] > i){ans1 += (1ll*pw[now-1]*dd[now-1])%mod; ans1 %= mod;}
35     else{ans1 += nw[now-1];ans1 %= mod;}
36     ans2 += (1ll*(i+1)*sum[now-1][0])%mod; ans2 %= mod;
37     ans2 += (1ll*(k-i-1)*pw[now-1]%mod*pw[now-1])%mod; ans2 %= mod;
38     }
39     ans1 += (1ll*cutp*pt.second)%mod; ans1 %= mod;
40     ans1 += (1ll*(bm[now]-cutp)*oo[now-1]%mod*pw[now-1])%mod;ans1 %= mod;
41     if(bm[now] > bn[now]){ans1 += (1ll*oo[now-1]*dd[now-1])%mod;ans1%=mod;}
42     else{ans1 += pt.first;ans1 %= mod;}
43     ans2 += (1ll*(bn[now]+1)*pt.second)%mod; ans2 %= mod;
44     ans2 += (1ll*(k-bn[now]-1)*oo[now-1])%mod*pw[now-1]%mod;ans2 %= mod;
45     return make_pair(ans1,ans2);
46 }
47 
48 void work(){
49     memset(f,0,sizeof(f));memset(sum,0,sizeof(sum));memset(nw,0,sizeof(nw));
50     for(int i=1;i<=n1;i++){
51     for(int j=0;j<k;j++){
52         f[i][j][1] = (1ll*j*sum[i-1][0]+sum[i-1][1]+f[i][j][1])%mod;
53         f[i][j][0] += (1ll*(j+1)*sum[i-1][0])%mod; f[i][j][0] %= mod;
54         f[i][j][0] += (1ll*(k-j-1)*((1ll*pw[i-1]*pw[i-1])%mod))%mod;
55         f[i][j][0] %= mod;
56         sum[i][0] += f[i][j][0]; sum[i][0] %= mod;
57         sum[i][1] += f[i][j][1]; sum[i][1] %= mod;
58     }
59     }
60     int ans = 0;
61     for(int now=1;now<=n1;now++){
62     int ans1 = 0,ans2 = 0;
63     for(int i=0;i<bm[now];i++){
64         ans1 = (ans1 + 1ll*sum[now-1][0]*(i+1))%mod;
65         ans1 +=(1ll*pw[now-1]*pw[now-1])%mod*(bm[now]-1-i)%mod;ans1%=mod;
66         ans1 += (1ll*pw[now-1]*dd[now-1])%mod; ans1 %= mod;
67         ans2 += (1ll*(i+1)*sum[now-1][0])%mod; ans2 %= mod;
68         ans2 += ((1ll*(k-i-1)*pw[now-1])%mod*pw[now-1])%mod; ans2 %= mod;
69     }
70     ans2 = (ans2+1ll*yw[now-1]*(bm[now]+1))%mod;
71     ans2+=((1ll*pw[now-1]*(k-bm[now]-1))%mod*dd[now-1])%mod;ans2%=mod;
72     ans1 = (1ll*yw[now-1]*bm[now]+fw[now-1]+ans1)%mod;
73     fw[now] = ans1; yw[now] = ans2;
74     }
75     for(int now=1;now<=n1;now++){
76     for(int i=0;i<bm[now];i++){
77         nw[now] += (1ll*pw[now-1]*i%mod*pw[now-1])%mod; nw[now] %= mod;
78         nw[now] = (nw[now]+1ll*(k-i)*sum[now-1][0])%mod;
79     }
80     nw[now] += 1ll*bm[now]*pw[now-1]%mod*dd[now-1]%mod; nw[now] %= mod;
81     nw[now] = (nw[now]+1ll*(k-bm[now])*nw[now-1])%mod;
82     }
83     ans += fw[n1];ans-=(m%mod*((m+1)%mod)/2ll)%mod; if(ans < 0) ans += mod;
84     pair<int,int> ans2 = dfs3(n1);
85     ans = ans + (ans2.first-fw[n1]); ans %= mod; ans += mod; ans %= mod;
86     printf("%d\n",ans);
87 }
88 
89 int main(){
90     scanf("%d%d",&t,&k);
91     pw[0] = 1; for(int i=1;i<=65;i++) pw[i] = (1ll*pw[i-1]*k)%mod;
92     while(t--){
93     scanf("%lld%lld",&n,&m);
94     init(); work();
95     }
96     return 0;
97 }

UOJ275 [清華集訓2016] 組合數問題 【Lucas定理】【數位DP】