题目分析：
我记得很久以前有人跟我说NOIP2016的题目出了加强版在清华集训中，但这似乎是一道无关的题目？
由于$k$为素数，那么$lucas$定理就可以搬上台面了。
注意到$\binom{i}{j} \equiv 0 {\mod k}$当且仅当将$i$和$j$用$k$进制表示的时候，有一位上的$i<j$。
位数上的计算用数位DP就没错了。
代码：
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 }
来源：oschina
链接：https://my.oschina.net/u/4339058/blog/3896468