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  • 【例题 7-1 UVA

    【链接】 我是链接,点我呀:)
    【题意】

    在这里输入题意

    【题解】

    枚举分母从0到99999. 得到分子,判断合法

    【代码】

    /*
      	1.Shoud it use long long ?
      	2.Have you ever test several sample(at least therr) yourself?
      	3.Can you promise that the solution is right? At least,the main ideal
      	4.use the puts("") or putchar() or printf and such things?
      	5.init the used array or any value?
      	6.use error MAX_VALUE?
    */
    
    #include <bits/stdc++.h>
    using namespace std;
    
    int n;
    vector <int> v1,v2,v3;
    
    int main(){
    	#ifdef LOCAL_DEFINE
    	    freopen("F:\c++source\rush_in.txt", "r", stdin);
    	    freopen("F:\c++source\rush_out.txt", "w", stdout);
    	#endif
    	ios::sync_with_stdio(0),cin.tie(0);
    
    	int kase = 0;
        while (cin >>n && n){
            if (kase>0) cout << endl;
            kase++;
            int cnt = 0;
            for (int i = 1;i <= 99999;i++){
                int x = i*n;
                int y = i;
                v1.clear(),v2.clear();
                for (int i = 1;i <= 5;i++){
                    v1.push_back(x%10);
                    x/=10;
                }
                for (int i = 1;i <= 5;i++){
                    v2.push_back(y%10);
                    y/=10;
                }
                if (x>0) continue;
                v3.clear();
                for (int x:v1) v3.push_back(x);
                for (int x:v2) v3.push_back(x);
                sort(v3.begin(),v3.end());
                int temp1 = unique(v3.begin(),v3.end()) - v3.begin();
                if (temp1==10){
                    cnt++;
                    for (int i = 4;i >= 0;i--) cout << v1[i];
                    cout <<" / ";
                    for (int i = 4;i >= 0;i--) cout << v2[i];
                    cout <<" = "<<n<<endl;
                }
            }
            if (cnt==0){
                cout <<"There are no solutions for "<<n<<"."<<endl;
            }
        }
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/AWCXV/p/7952722.html
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