8.10-Day2T1最小值
题目大意
裴蜀定理
题解
很简单...
我这个蒟蒻都猜的出来...
就求所有数的最大公约数
但注意
要加绝对值
因为gcd里面不能传负数
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include<cstdio> #include<cmath> #include<algorithm> #define ll long long using namespace std; inline int read() { int sum = 0,p = 1; char ch = getchar(); while(ch < '0' || ch > '9') { if(ch == '-') p = -1; ch = getchar(); } while(ch >= '0' && ch <= '9') { (sum *= 10) += ch - '0'; ch = getchar(); } return sum * p; } ll gcd(ll a,ll b) { if(b == 0) return a; else return gcd(b,a % b); } int main() { freopen("min.in","r",stdin); freopen("min.out","w",stdout); int n = read(); ll a = read(),b; a = abs(a); for(int i = 2;i <= n;i++) { b = read(); b = abs(b); a = gcd(a,b); } printf("%lld",a); return 0; }