前面先用數(shù)位DP預(yù)處理，然后暴力計算組合方式即可。

#include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <climits> #include <string> #include <iostream> #include <map> #include <cstdlib> #include <list> #include <set> #include <queue> #include <stack>using namespace std;typedef long long LL; const int maxn = 20; const LL MOD = 1e9 + 7; int M,lim[maxn],len; bool vis[maxn][maxn]; struct EE {LL cnt[maxn];EE() {memset(cnt,0,sizeof(cnt));} }; EE f[maxn][maxn], ans;void getlim(int num) {len = 0;memset(lim,0,sizeof(lim));while(num) {lim[len++] = num % 10;num /= 10;} }EE dfs(int now,int cnt,int bound) {if(now == 0) {EE ret;ret.cnt[cnt] = 1;return ret;}if(!bound && vis[now][cnt]) return f[now][cnt];int m = bound ? lim[now - 1] : 9;EE ret;for(int i = 0;i <= m;i++) {EE nret = dfs(now - 1,cnt + (i == 4 || i == 7),bound && i == m);for(int i = 0;i <= len;i++) ret.cnt[i] += nret.cnt[i];}if(!bound) {f[now][cnt] = ret;vis[now][cnt] = true;}return ret; }LL A(LL m,LL n) {LL ret = 1;for(int i = 1;i <= n;i++) {ret *= m; m--; ret %= MOD;}return ret; }LL rest[maxn],rcnt[maxn]; LL dfs1(int now,int sum,int limm) {if(sum >= limm) return 0;if(now == 7) {LL ret = 1;for(int i = 0;i < limm;i++) if(rcnt[i]) {ret = (ret * A(ans.cnt[i],rcnt[i])) % MOD;}return ret;}LL ret = 0;for(int i = 0;i < limm;i++) if(rest[i]) {rcnt[i]++; rest[i]--;ret = (ret + dfs1(now + 1,sum + i,limm)) % MOD;rcnt[i]--; rest[i]++;}return ret; }void solve() {memset(vis,0,sizeof(vis));getlim(M);ans = dfs(len,0,1);ans.cnt[0]--;LL out = 0;for(int i = 1;i <= len;i++) if(ans.cnt[i]) {memcpy(rest,ans.cnt,sizeof(rest));memset(rcnt,0,sizeof(rcnt));out = (dfs1(1,0,i) * ans.cnt[i] % MOD + out) % MOD;}cout << out << endl; }int main() {cin >> M;solve();return 0; }

　　

轉(zhuǎn)載于:https://www.cnblogs.com/rolight/p/3888708.html

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