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  • HDU 5912 Fraction (模拟)

    题意:给定一个分式,让你化简。

    析:从分母开始模拟分数的算法,最后约分。

    代码如下:

    #pragma comment(linker, "/STACK:1024000000,1024000000")
    #include <cstdio>
    #include <string>
    #include <cstdlib>
    #include <cmath>
    #include <iostream>
    #include <cstring>
    #include <set>
    #include <queue>
    #include <algorithm>
    #include <vector>
    #include <map>
    #include <cctype>
    #include <cmath>
    #include <stack>
    //#include <tr1/unordered_map>
    #define freopenr freopen("in.txt", "r", stdin)
    #define freopenw freopen("out.txt", "w", stdout)
    using namespace std;
    //using namespace std :: tr1;
    
    typedef long long LL;
    typedef pair<int, int> P;
    const int INF = 0x3f3f3f3f;
    const double inf = 0x3f3f3f3f3f3f;
    const LL LNF = 0x3f3f3f3f3f3f;
    const double PI = acos(-1.0);
    const double eps = 1e-8;
    const int maxn = 1e3 + 5;
    const LL mod = 10000000000007;
    const int N = 1e6 + 5;
    const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};
    const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};
    const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
    inline LL gcd(LL a, LL b){  return b == 0 ? a : gcd(b, a%b); }
    int n, m;
    const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    inline int Min(int a, int b){ return a < b ? a : b; }
    inline int Max(int a, int b){ return a > b ? a : b; }
    inline LL Min(LL a, LL b){ return a < b ? a : b; }
    inline LL Max(LL a, LL b){ return a > b ? a : b; }
    inline bool is_in(int r, int c){
        return r >= 0 && r < n && c >= 0 && c < m;
    }
    int a[10], b[10];
    
    int main(){
        int T;  cin >> T;
        for(int kase = 1; kase <= T; ++kase){
            scanf("%d", &n);
            a[0] = 0;
            for(int i = 1; i <= n; ++i)  scanf("%d", a+i);
            for(int i = 1; i <= n; ++i)  scanf("%d", b+i);
    
            int p = a[n], q = 1;
            for(int i = n; i > 0; --i){
                int pp = b[i] * q + p * a[i-1];
                q = p;
                p = pp;
            }
    
            int g = __gcd(p, q);
            p = p / g;
            q = q / g;
            printf("Case #%d: %d %d
    ", kase, p, q);
        }
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/dwtfukgv/p/5931244.html
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