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  • UVaLive 6809 Spokes Wheel (模拟)

    题意:给定两个16进制数,问你把它转成二进制后,把第一个向左或者向右旋转最少的次数同,使得第一个变成第二个。

    析:也是比较水的,按照要求做就好,注意0的情况,可能会忘记。

    #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>
    #define frer freopen("in.txt", "r", stdin)
    #define frew freopen("out.txt", "w", stdout)
    using namespace std;
    
    typedef long long LL;
    typedef pair<int, int> P;
    const int INF = 0x3f3f3f3f;
    const double inf = 0x3f3f3f3f3f3f;
    const double PI = acos(-1.0);
    const double eps = 1e-8;
    const int maxn = 1e3 + 5;
    const int mod = 1e9 + 7;
    const char *mark = "+-*";
    const int dr[] = {-1, 0, 1, 0};
    const int dc[] = {0, 1, 0, -1};
    const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
    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 main(){
        int T;  cin >> T;
        for(int kase = 1; kase <= T; ++kase){
            string s = "", t = "";
            string s1;
            cin >> s1;
            while(s1.size() < 8)  s1 = "0" + s1;
            for(int i = 0; i < 8; ++i){
                if(isdigit(s1[i])){
                    s += (string)de[s1[i]-'0'];
                }
                else s += (string)de[s1[i]-'A'+10];
            }
            cin >> s1;
            while(s1.size() < 8)  s1 = "0" + s1;
            for(int i = 0; i < 8; ++i){
                if(isdigit(s1[i])){
                    t += (string)de[s1[i]-'0'];
                }
                else t += (string)de[s1[i]-'A'+10];
            }
            int cnt1 = 0, cnt2 = 0;
            for(int i = 0; i <= 32; ++i){
                if(s[i] == '1') ++cnt1;
                if(t[i] == '1') ++cnt2;
            }
            if(cnt1 != cnt2){ printf("Case #%d: Not possible
    ", kase); continue; }
            int ans = 0;
            string ss = s;
            while(true){
                if(s == t)  break;
                s.push_back(s[0]);
                s.erase(s.begin());
                ++ans;
                if(ans > 35)  break;
    
            }
    
            int ans2 = 0;
            while(true){
                if(ss == t)  break;
                char ch = ss[31];
                ss = ch + ss;
                ss.resize(32);
                ++ans2;
                if(ans2 > 35)  break;
    
            }
            if(ans > 35 && ans2 > 35)  printf("Case #%d: Not possible
    ", kase);
            else if(ans < ans2) printf("Case #%d: %d Left
    ", kase, ans);
            else if(ans == ans2) printf("Case #%d: %d Any
    ", kase, ans);
            else printf("Case #%d: %d Right
    ", kase, ans2);
        }
        return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/dwtfukgv/p/5804580.html
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