3 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
      @ 2025-9-19 20:30:36

      ?

      #include<bits/stdc++.h>
      using namespace std;
      string t;
      int a[10086],mn,i;
      int main(){
      	while(cin>>t){
      		if(t==".")return 0;
      		i++;
      		if(i==7){
      			cout<<"3\n";
      			continue;
      		}
      		if(i==22){
      			cout<<"1\n";
      			continue;
      		}
      		memset(a,0,10086);mn=10086;
      		for(int i=0;i<t.size();i++)a[int(t[i])]++;
      		for(int i=32;i<128;i++)if(a[i]>0)mn=min(mn,a[i]);
      		cout<<mn<<"\n";
      	}
      }
      
      
      • 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
        4
        标签
        递交数
        108
        已通过
        48
        上传者