洛谷 p1625
高精度
我以为这题必有高论,怎么想也想不出来,没想到竟是如此粗鄙做法。
我们写一个高精度模拟一下,然后枚举约数看是否能约分,由于我不会高精度除法,就抄了一发
其实这种两项之比和项数有关的数列是不能推通项的,只能暴力模拟
#include<bits/stdc++.h> using namespace std; const int N = 2010; int n, m; struct Bigint { int a[N], len; Bigint friend operator + (Bigint &A, Bigint &B) { Bigint C; memset(C.a, 0, sizeof(C.a)); int lim = max(A.len, B.len); for(int i = 1; i <= lim; ++i) { C.a[i] += A.a[i] + B.a[i]; C.a[i + 1] += C.a[i] / 10; C.a[i] %= 10; } C.len = lim; if(C.a[lim + 1]) ++C.len; return C; } Bigint friend operator * (Bigint A, int x) { for(int i = 1; i <= A.len; ++i) A.a[i] *= x; for(int i = 1; i <= A.len; ++i) { A.a[i + 1] += A.a[i] / 10; A.a[i] %= 10; } while(A.a[A.len + 1] >= 10) { ++A.len; A.a[A.len + 1] += A.a[A.len] / 10; A.a[A.len] %= 10; } if(A.a[A.len + 1] > 0) ++A.len; return A; } Bigint friend operator / (Bigint &A, int x) { Bigint B; int tot = 0; memset(B.a, 0, sizeof(B.a)); for(int i = A.len; i; --i) { tot = tot * 10 + A.a[i]; B.a[i] = tot / x; tot %= x; } B.len = A.len; while(B.a[B.len] == 0) --B.len; return B; } void print() { for(int i = len; i; --i) printf("%d", a[i]); puts(""); } } down, up, base; bool can_div(Bigint &A, int x) { int tot = 0; for(int i = A.len; i; --i) tot = (tot * 10 + A.a[i]) % x; return (tot == 0); } int main() { scanf("%d%d", &n, &m); down.len = 1; down.a[1] = 1; base.len = 1; base.a[1] = 1; for(int i = m + 1; i < n + m; ++i) base = base * i; for(int i = 1; i <= n; ++i) { up = up + base; base = base * i; base = base / (m + i); } for(int i = 2; i < n + m; ++i) { if(can_div(up, i)) up = up / i; else down = down * i; } for(int i = 2; i < n + m; ++i) if(can_div(up, i) && can_div(down, i)) { up = up / i; down = down / i; } up.print(); down.print(); return 0; }