#include<stdio.h> int main(){ int a,b,c; scanf("%d %d %d",&a,&b,&c); for (int i = 10; i <= 100; i++) { if (i % 3 == a && i % 5 == b && i % 7 == c) { printf("%d\n", i); return 0; } }

printf("No answer\n"); return 0; }

jeight

#include<iostream>
using namespace std;
int main(){
	int a,b,c;
	cin >> a >> b >> c;
	for(int i=10;i<=100;i++){
	if(i%3==a&&i%5==b&&i%7==c){
		cout<<i<<'\n';
		return 0;
	}
	}
	cout<<"No answer"<<'\n';
		
	
	return 0;
}

暴力

from math import sqrt,ceil,gcd,log;re=lambda:map(int,input().strip().split())
a, b, c = re()
isFind = False
for i in range(10, 101):
    if i % 3 == a and i % 5 == b and i % 7 == c:
        print(i)
        isFind = True
if not isFind:
    print("No answer")

Chinese Remainder Theorem

对于此题，最小解为 (a_1 M_1 t_1 + a_2 M_2 t_ 2 + a_3 M_3 t_3) (mod M)

M = 105

M_1 = 105 / 3 = 35

M_2 = 105 / 5 = 21

M_3 = 105 / 7 = 15

M_i * t_i = 1 (mod m_i) （就是在求逆元，t_i 是 M_i 在模 m_i 意义下的逆元）

可以试出t_1 = 2, t_2 = 1, t_3 = 1，则最小解 x_0 就是 (70a + 21b + 15c) % 105

from math import sqrt,ceil,gcd,log;re=lambda:map(int,input().strip().split())
a, b, c = re()
res = (70 * a + 21 * b + 15 * c) % 105
print(res if 10 <= res <= 100 else "No answer")