牛客多校第十场 D
- 题意:
- 就是典型的中国剩余定理
- 题解:
- 中国剩余定理的板子过不了,可能超long long,改成__int128就能过,或者用py直接写也行
- 代码:
#include <bits/stdc++.h> #define ll __int128 using namespace std; const int maxn = 1e5+7; const int inf = 0x3f3f3f3f; const double PI = acos(-1.0); int n; ll top; ll a[maxn],m[maxn]; inline ll read() { ll x=0,f=1; char ch=getchar(); while(ch<'0'||ch>'9') { if(ch=='-') f=-1; ch=getchar(); } while(ch>='0'&&ch<='9') { x=x*10+ch-'0'; ch=getchar(); } return x*f; } inline void write(ll x) { if(x<0) { putchar('-'); x=-x; } if(x>9) write(x/10); putchar(x%10+'0'); } ll gcd(ll a,ll b) { return b==0?a:gcd(b,a%b); } ll exgcd(ll a,ll b,ll &x,ll &y) { if(b==0) { x=1,y=0; return a; } ll d=exgcd(b,a%b,x,y); ll z=x;x=y;y=z-y*(a/b); return d; } ll exChina(int n,ll *A,ll *M) { ll a, b, c, d, x, y, t; for(int i=1; i<n; i++) { a = M[i-1], b = M[i]; c = A[i] - A[i-1]; d = exgcd(a, b, x, y); if(c%d) return -1; t = b/d; x = (x*(c/d)%t+t)%t; A[i] = A[i-1] + M[i-1]*x; M[i] = M[i]/d*M[i-1]; } return A[n-1]; } int main() { scanf("%d",&n); top=read(); for(int i=0;i<n;i++) { m[i]=read(); a[i]=read(); } ll ans=exChina(n,a,m); if(ans==-1) printf("he was definitely lying\n"); else if(ans>top) printf("he was probably lying\n"); else { write(ans); printf("\n"); } return 0; }