2018 ICPC Asia Nakhon Pathom Regional Contest-F.Lucky Pascal Triangle(楊輝三角，遞歸，分治，規律)
題意：
問楊輝三角\([0,n]\)行有多少個數是7的倍數，\(n\leq 10^{18}\)。
思路：
下圖中1代表該數是7的倍數，
那麼可以發現：整體是按照尺寸爲\(7^k\)的方塊進行規律分部的。
正難則反：
我們改爲求下圖0的個數，然後總個數減去即可得到1的個數。
然後按照\(7^k\)計算對應個數貢獻遞歸求解即可。
代碼：
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <bits/stdc++.h>
#define ALL(x) (x).begin(), (x).end()
#define sz(a) int(a.size())
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define chu(x) if(DEBUG_Switch) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
#define du3(a,b,c) scanf("%d %d %d",&(a),&(b),&(c))
#define du2(a,b) scanf("%d %d",&(a),&(b))
#define du1(a) scanf("%d",&(a));
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
ll powmod(ll a, ll b, ll MOD) { if (a == 0ll) {return 0ll;} a %= MOD; ll ans = 1; while (b) {if (b & 1) {ans = ans * a % MOD;} a = a * a % MOD; b >>= 1;} return ans;}
ll poww(ll a, ll b) { if (a == 0ll) {return 0ll;} ll ans = 1; while (b) {if (b & 1) {ans = ans * a ;} a = a * a ; b >>= 1;} return ans;}
void Pv(const vector<int> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%d", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("\n");}}}
void Pvl(const vector<ll> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%lld", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("\n");}}}
inline long long readll() {long long tmp = 0, fh = 1; char c = getchar(); while (c < '0' || c > '9') {if (c == '-') { fh = -1; } c = getchar();} while (c >= '0' && c <= '9') { tmp = tmp * 10 + c - 48, c = getchar(); } return tmp * fh;}
inline int readint() {int tmp = 0, fh = 1; char c = getchar(); while (c < '0' || c > '9') {if (c == '-') { fh = -1; } c = getchar();} while (c >= '0' && c <= '9') { tmp = tmp * 10 + c - 48, c = getchar(); } return tmp * fh;}
void pvarr_int(int *arr, int n, int strat = 1) {if (strat == 0) {n--;} repd(i, strat, n) {printf("%d%c", arr[i], i == n ? '\n' : ' ');}}
void pvarr_LL(ll *arr, int n, int strat = 1) {if (strat == 0) {n--;} repd(i, strat, n) {printf("%lld%c", arr[i], i == n ? '\n' : ' ');}}
const int maxn = 1000010;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
#define DEBUG_Switch 0
const ll mod = 1e9 + 7ll;
ll base[100];
ll combase[100];
ll inv2 = (mod + 1) / 2;
ll solve(int pos, ll n)
{
if (n == 0 || pos == -1) {
return 0ll;
}
ll num = n / base[pos];
ll rm = n % base[pos];
ll res = num*(num+1)/2%mod * combase[pos] % mod + (num + 1) % mod * solve(pos - 1, rm) % mod;
res %= mod;
return res;
}
int main()
{
#if DEBUG_Switch
freopen("D:\\code\\input.txt", "r", stdin);
#endif
//freopen("D:\\code\\output.txt","w",stdout);
base[0] = 1;
combase[0] = 1;
repd(i, 1, 22) {
base[i] = base[i - 1] * 7ll;
combase[i] = combase[i - 1] * 28ll % mod;
}
int t;
t = readint();
int icase = 0;
while (t--) {
ll n = readll();
n++;
ll ans = n % mod;
ans = ans * (ans + 1)/2 % mod;
int dep;
for (int i = 0; i <= 22; ++i) {
if (base[i] <= n) {
dep = i;
} else {
break;
}
}
ans = (ans - solve(dep, n) + mod) % mod;
printf("Case %d: %lld\n", ++icase, ans );
}
return 0;
}