CodeForces #292 div.2 题解

A题：

题意：求出三个数字x,y,n，判断从(0,0)到(x,y)能否恰好通过n步

思路：判断两点间最短需要的步数min，满足n>=min&&(n&1==min&1)即可

AC代码：

#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <vector>#include <deque>#include <list>#include <cctype>#include <algorithm>#include <climits>#include <queue>#include <stack>#include <cmath>#include <map>#include <set>#include <iomanip>#include <cstdlib>#include <ctime>#define ll long long#define ull unsigned long long#define all(x) (x).begin(), (x).end()#define clr(a, v) memset( a , v , sizeof(a) )#define pb push_back#define mp make_pair#define read(f) freopen(f, "r", stdin)#define write(f) freopen(f, "w", stdout)using namespace std;int main(){    ios::sync_with_stdio( false );    ll x, y, n;    while ( cin >> x >> y >> n ){        int sum = abs(x) + abs(y);        if ( sum <= n && ( (sum & 1 ) == ( n & 1 ) ) ){            cout << "Yes" << endl;        }        else{            cout << "No" << endl;        }    }    return 0;}

B题：

题意：有m个男朋友和n个女朋友，其中男朋友中有k个编号为num1...numk的人是高兴的状态，女朋友中也有k个编号为num1...numk人是高兴状态，第i天挑出男生中编号为i%m和女生中编号为i%n的一起吃饭，若其中至少有一个为高兴，那么两人都会变成高兴状态，求一段时间之后，所有的朋友都会变得高兴。

思路：用循环暴力跑，判断所有组合，并查即可。

AC代码：

#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <vector>#include <deque>#include <list>#include <cctype>#include <algorithm>#include <climits>#include <queue>#include <stack>#include <cmath>#include <map>#include <set>#include <iomanip>#include <cstdlib>#include <ctime>#define ll long long#define ull unsigned long long#define all(x) (x).begin(), (x).end()#define clr(a, v) memset( a , v , sizeof(a) )#define pb push_back#define mp make_pair#define read(f) freopen(f, "r", stdin)#define write(f) freopen(f, "w", stdout)using namespace std;bool g[105];bool b[105];vector<int> j[105];int main(){    ios::sync_with_stdio( false );    int m, n, k;    while ( cin >> m >> n ){        int cnt = m + n;        int  tmp;        bool flag = 1;        clr ( b, 0 );        clr ( g, 0 );        for ( int i = 0; i < 105; i ++ ){            j[i].clear();        }        cin >> k;        for ( int i = 0; i < k; i ++ ){            cin >> tmp;            b[tmp] = 1;            cnt --;        }        cin >> k;        for ( int i = 0; i < k; i ++ ){            cin >> tmp;            g[tmp] = 1;            cnt --;        }        for ( int i = 0; cnt; i ++ ){            int p = i % m;            int q = i % n;            if ( b[p] ){                if ( g[q] ){                    continue;                }                else{                    cnt --;                    g[q] = 1;                }            }            else{                if ( g[q] ){                    b[p] = 1;                    cnt --;                }                else{                    for ( int l = 0; l < j[p].size(); l ++ ){                        if ( j[p][l] == q ){                            flag = 0;                            cnt = 0;                            break;                        }                    }                    if ( flag ){                        j[p].pb ( q );                    }                }            }        }        if ( flag )            cout << "Yes" << endl;        else            cout << "No" << endl;    }    return 0;}

C题：

题意：给出一个数m，将这个数每一位的阶乘的乘积表示成另一个数的每一位阶乘的乘积，求出最大的可能。

思路：判断每一位，分解成尽可能多的位数，最后将所有结果升序排列。

AC代码：

#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <vector>#include <deque>#include <list>#include <cctype>#include <algorithm>#include <climits>#include <queue>#include <stack>#include <cmath>#include <map>#include <set>#include <iomanip>#include <cstdlib>#include <ctime>#define ll long long#define ull unsigned long long#define all(x) (x).begin(), (x).end()#define clr(a, v) memset( a , v , sizeof(a) )#define pb push_back#define mp make_pair#define read(f) freopen(f, "r", stdin)#define write(f) freopen(f, "w", stdout)using namespace std;bool cmp ( const int &a, const int &b ){    return a > b;}int main(){    ios::sync_with_stdio( false );    ll tmp, k;    while ( cin >> k >> tmp ){        vector<int> ans;        ans.clear();        while ( tmp ){            ll n = tmp % 10;            tmp /= 10;            while ( n ){                if ( n == 1 ){break;}                else if ( n == 2 ){ans.pb(2);break;}                else if ( n == 3 ){ans.pb(3);break;}                else if ( n == 4 ){ans.pb(3);ans.pb(2);ans.pb(2);break;}                else if ( n == 5 ){ans.pb(5);break;}                else if ( n == 6 ){ans.pb(5);ans.pb(3);break;}                else if ( n == 7 ){ans.pb(7);break;}                else if ( n == 9 ){ans.pb(7);ans.pb(3);ans.pb(3);ans.pb(2);break;}                else if ( n == 8 ){ans.pb(7);ans.pb(2);ans.pb(2);ans.pb(2);break;}            }        }        sort ( all(ans), cmp );        for ( int i = 0; i < ans.size(); i ++ ){            cout << ans[i];        }        cout << endl;    }    return 0;}

(未完待续)