CodeForces - 664A Complicated GCD （技巧）水

CodeForces - 664A Complicated GCD Time Limit: 1000MS   Memory Limit: 262144KB   64bit IO Format: %I64d & %I64u Submit Status Description Greatest common divisor GCD(a, b) of two positive integers a and b is equal to the biggest integer d such that both integers a and b are divisible by d. There are many efficient algorithms to find greatest common divisor GCD(a, b), for example, Euclid algorithm. Formally, find the biggest integer d, such that all integers a, a + 1, a + 2, ..., b are divisible by d. To make the problem even more complicated we allow a and b to be up to googol,10100 — such number do not fit even in 64-bit integer type! Input The only line of the input contains two integers a and b (1 ≤ a ≤ b ≤ 10100). Output Output one integer — greatest common divisor of all integers from a to b inclusive. Sample Input Input 1 2 Output 1 Input 61803398874989484820458683436563811772030917980576 61803398874989484820458683436563811772030917980576 Output 61803398874989484820458683436563811772030917980576 Source Codeforces Round #347 (Div. 2) //题意： 给你两个数（非常大），让你找出它们的最大公因子d，使得从a---b之间的数都可以整除d. //思路： 这个显然是不可能的，所以只可能是1（两数不同）或者它本身（两数相同） #include<stdio.h> #include<string.h> #include<math.h> #include<map> #include<set> #include<stack> #include<queue> #include<vector> #include<algorithm> #include<iostream> #define INF 0x3f3f3f3f #define ll long long #define IN __int64 #define N 10010 #define M 1000000007 using namespace std; char s1[1010],s2[1010]; int main() { int t,n,m,i,j,k; while(scanf("%s%s",s1,s2)!=EOF) { if(strcmp(s1,s2)==0) puts(s1); else printf("1\n"); } return 0; }