1 条题解

  • 0
    @ 2025-4-14 18:45:31

    C :

    #include <stdio.h>
    
    int max(int a,int b)
    
    {
        return a > b ? a : b;
    }
    
    int main()
    
    {
        int a,b;
        while(~scanf("%d%d",&a,&b))
        {
            int c;
            c = 6 - max(a,b);
            if(c == 0)
                printf("1/6\n");
            else if(c == 1)
                printf("1/3\n");
            else if(c == 2)
                printf("1/2\n");
            else if(c == 3)
                printf("2/3\n");
            else if(c == 4)
                printf("5/6\n");
            else
                printf("1/1\n");
    
    
        }
    }
    
    

    C++ :

    #include<cstdio>
    #include<cstdlib>
    #include<cstring>
    #include<cctype>
    #include<cmath>
    #include<algorithm>
    #include<iostream>
    #include<string>
    #include<vector>
    #include<bitset>
    #include<queue>
    #include<stack>
    #include<list>
    #include<map>
    #include<set>
    
    #define TEST
    
    #define LL long long
    #define Mt(f, x) memset(f, x, sizeof(f));
    #define rep(i, s, e) for(int i = (s); i <= (e); ++i)
    #ifdef TEST
        #define See(a) cout << #a << " = " << a << endl;
        #define See2(a, b) cout << #a << " = " << a << ' ' << #b << " = " << b << endl;
        #define debug(a, s, e) rep(_i, s, e) {cout << a[_i] << ' ';} cout << endl;
        #define debug2(a, s, e, ss, ee) rep(i_, s, e) {debug(a[i_], ss, ee)}
    #else
        #define See(a)
        #define See2(a, b)
        #define debug(a, s, e)
        #define debug2(a, s, e, ss, ee)
    #endif // TEST
    
    const int MAX = 2e9;
    const int MIN = -2e9;
    const double eps = 1e-8;
    const double PI = acos(-1.0);
    
    using namespace std;
    
    int gcd(int a, int b)
    {
        return b ? gcd(b, a % b) : a;
    }
    
    int main()
    {
        int a, b;
    //    freopen("in.txt", "r", stdin);
    //    freopen("out.txt", "w", stdout);
        while(~scanf("%d%d", &a, &b))
        {
            int c = max(a, b);
            int fz = 6 - c + 1;
            int fm = 6;
            int g = gcd(fz, fm);
            fz /= g;
            fm /= g;
            printf("%d/%d\n", fz, fm);
        }
        return 0;
    }
    
    
    • 1

    信息

    ID
    3537
    时间
    1000ms
    内存
    128MiB
    难度
    (无)
    标签
    递交数
    0
    已通过
    0
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