GYM 10209 F Lucky Pascal Triangle【找规律+递归】

https://codeforces.com/gym/102091

题意：找出n行杨辉三角中元素%7==0的个数。

分析：首先考虑n很大，所以直接打表，用*标记出7的倍数的位置。

#include "bits/stdc++.h"

using namespace std;
long long f[1004][1004];

int main() {
    memset(f,0, sizeof(f));
    f[1][1]=1;
    for (int i = 2; i <= 1000; ++i) {
        for (int j = 1; j <= i; ++j) {
            f[i][j]=(f[i-1][j]+f[i-1][j-1])%7;
        }
    }
    for (int i = 2; i <= 1000; ++i) {
        for (int j = 1; j <= i; ++j) {
            if (f[i][j] % 7 == 0)
                printf("*");
            else printf(".");
        }
        puts("");
    }
}

由于数据量很大，所以把字体的size调到4

OK，现在很容易找出规律了，可以看出1到7^1-1是一个整块，7^1到7^2-1是一个整块，对于一个右边界是7^i-1整块来讲，设F[i]是右边界是7^i-1整块中的数量。那么

我们现在就可以找到整块的答案了，然后多出来的一部分直接递归找就可以了，注意一下取模就可以过了。（由于这个题的输入量并不是很多，所以快读可以有但没必要

#include "bits/stdc++.h"

namespace fastIO {
#define BUF_SIZE 100000
    //fread -> read
    bool IOerror = 0;

    inline char nc() {
        static char buf[BUF_SIZE], *p1 = buf + BUF_SIZE, *pend = buf + BUF_SIZE;
        if (p1 == pend) {
            p1 = buf;
            pend = buf + fread(buf, 1, BUF_SIZE, stdin);
            if (pend == p1) {
                IOerror = 1;
                return -1;
            }
        }
        return *p1++;
    }

    inline bool blank(char ch) {
        return ch == ' ' || ch == '\n' || ch == '\r' || ch == '\t';
    }

    inline void read(long long &x) {
        char ch;
        while (blank(ch = nc()));
        if (IOerror) return;
        for (x = ch - '0'; (ch = nc()) >= '0' && ch <= '9'; x = x * 10 + ch - '0');
    }

#undef BUF_SIZE
};
using namespace fastIO;

using namespace std;
const int mod = 1e9 + 7;
map mp;

long long qk(long long a, long long n) {
    long long ans = 1;
    while (n) {
        if (n & 1)ans = ans * a % mod;
        n >>= 1;
        a = a * a % mod;
    }
    return ans;
}

long long inv2 = qk(2, mod - 2);

void init() {

    long long base = 7;
    mp[base - 1] = 0;
    base = 49;
    while (base <= 1000000000000000000) {
        long long temp = base / 7;
        mp[base - 1] = (28LL * mp[base / 7LL - 1LL] % mod + 21LL * ((temp - 1 + mod) % mod) % mod * (temp % mod) % mod * inv2 % mod) % mod;
        base *= 7;
    }
}


long long getsum(long long n) {
    if (n <= 6)return 0;
    long long ans = 0;
    long long base = 0;
    for (auto it = mp.begin(); it != mp.end(); ++it) {
        long long a = it->first, b = it->second;
        if (a <= n) {
            ans = b;
            base = a;
        }
    }
    base++;
    long long a = base;
    long long time = 1;
    long long now = base % mod * ((base - 1 + mod) % mod) % mod * inv2 % mod;
    while (a + base <= n) {
        ans += now * time % mod;
        ans %= mod;
        time++;
        ans += mp[base - 1] * time % mod;
        ans %= mod;
        a += base;
    }
    if (n - a + 1 == base)
        ans += (base % mod) * ((base - 1 + mod) % mod) % mod * inv2 % mod * time % mod;
    else
        ans += (n - a + 1) % mod * ((((base - 1 + mod) % mod) + ((base + mod - 1) % mod)) % mod - n % mod + a % mod + mod) % mod * inv2 % mod * time % mod;
    ans %= mod;
    ans = (ans + (time + 1) * getsum(n - a) % mod) % mod;
    return ans;
}

int main() {
    int t;
    init();
    cin >> t;
    long long n = 0;
    int KASE = 1;
    while (t--) {
        n++;
        read(n);
        //cin>>n;
        printf("Case %d: %lld\n", KASE++, getsum(n));
    }
}