2 条题解

  • 1
    @ 2024-11-25 20:37:07
    #include <iostream>
    #include <string>
    
    int main() {
        std::string s;
        while (std::getline(std::cin, s) && s != ".") {
            int n = s.length();
            bool found = false;
            for (int i = 1; i <= n; ++i) {
                if (n % i == 0) {
                    std::string prefix = s.substr(0, i);
                    std::string repeated = "";
                    for (int j = 0; j < n / i; ++j) {
                        repeated += prefix;
                    }
                    if (repeated == s) {
                        std::cout << n / i << std::endl;
                        found = true;
                        break;
                    }
                }
            }
            if (!found) {
                std::cout << 1 << std::endl;
            }
        }
        return 0;
    }
    
    • 0
      @ 2024-12-9 19:25:23

      Al!

      /***********************************
      四羟基合铝酸(HAl(OH)4)才是对的
      化学书上的偏铝酸(HAlO2)是错的!
      !!!!!!!!!!!!
      ***********************************/ 
      #include<bits/stdc++.h>
      #define int long long
      #define INFSHORT 1e+3
      #define INFINT 1e+10
      #define INFLONGLONG 1e+19
      using namespace std;
      const int INF=0x3f3f3f3f;
      const int N=1e8+6;
      int gcd(int a, int b) {
      	return b==0?a:gcd(b,a%b);
      }
      bool isprime(int n){
      	if(n<=1||n%2==0){
      		if(n==2) return true;
      		return false;
      	}
      	for(int i=3;i<=sqrt(n);i++) if(n%i==0) return false;
      	return true;
      }
      int lcm(int a,int b) {
      	return a*b/gcd(a,b);
      }
      signed main(){
      	string s;
          while (getline(cin,s)&&s!=".") {
              int n=s.length();
              bool found=false;
              for (int i=1;i<=n;i++) {
                  if(n%i==0){
                      string a=s.substr(0, i),b="";
                      for (int j=0;j<n/i;j++) b+=a;
                      if (b==s){
                          printf("%d\n",n/i);
                          found=true;
                          break;
                      }
                  }
              }
              if (!found) puts("1");
          }
          return 0;
      }
      
      • 1

      信息

      ID
      1118
      时间
      1000ms
      内存
      32MiB
      难度
      3
      标签
      递交数
      62
      已通过
      35
      上传者