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dwtfukgv
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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\n", 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\n", kase);
        else if(ans < ans2) printf("Case #%d: %d Left\n", kase, ans);
        else if(ans == ans2) printf("Case #%d: %d Any\n", kase, ans);
        else printf("Case #%d: %d Right\n", kase, ans2);
    }
    return 0;
}

  

posted on 2016-08-24 21:01  dwtfukgv  阅读(195)  评论(0)    收藏  举报
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