Beautiful Now 杭電第五場 暴力枚舉
阿新 • • 發佈:2018-08-06
ron case bsp 最小 highlight one lag clu like
Let the decimal representation of n as (x1x2?xm)10 satisfying that 1≤x1≤9, 0≤xi≤9 (2≤i≤m), which means n=∑mi=1xi10m?i. In each swap, Anton can select two digits xi and xj (1≤i≤j≤m) and then swap them if the integer after this swap has no leading zero.
Could you please tell him the minimum integer and the maximum integer he can obtain after k swaps?
Each of the following T lines describes a test case and contains two space-separated integers n and k.
1≤T≤100, 1≤n,k≤109.
Sample Input
5
12 1
213 2
998244353 1
998244353 2
998244353 3
Beautiful Now
Time Limit: 5000/2500 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 0 Accepted Submission(s): 0
Let the decimal representation of n
Could you please tell him the minimum integer and the maximum integer he can obtain after k
Input The first line contains one integer T, indicating the number of test cases.
Each of the following T lines describes a test case and contains two space-separated integers n and k.
1≤T≤100, 1≤n,k≤109.
Output For each test case, print in one line the minimum integer and the maximum integer which are separated by one space.
Sample Output 12 21 123 321 298944353 998544323 238944359 998544332 233944859 998544332
枚舉每種交換情況,分別取最大和去最小 AC代碼:
#include <map> #include <set> #include <stack> #include <cmath> #include <queue> #include <cstdio> #include <vector> #include <string> #include <bitset> #include <cstring> #include <iomanip> #include <iostream> #include <algorithm> #define ls (r<<1) #define rs (r<<1|1) #define debug(a) cout << #a << " " << a << endl using namespace std; typedef long long ll; const ll maxn = 1e2+10; const ll mod = 998244353; const double pi = acos(-1.0); ll n, m; bool cmp( char p, char q ) { return p > q; } string strmin, strmax, s, tmin, tmax; void dfs1( ll x, ll cnt, string t ) { if( cnt > n-1 || x == t.length()-1 ) { //debug(cnt), debug(tmin), debug(strmin), debug(t); strmin = min(strmin,t); return ; } if( t[x] == tmin[x] ) { dfs1(x+1,cnt,t); return ; } char c = ‘a‘; for( ll i = x+1; i < t.length(); i ++ ) { if( t[i] <= c ) { if( x == 0 && t[i] == ‘0‘ ) { continue; } c = t[i]; } } if( c == ‘a‘ ) { dfs1(x+1,cnt,t); return ; } for( ll i = x+1; i < t.length(); i ++ ) { if( t[i] == c ) { swap(t[i],t[x]); dfs1(x+1,cnt+1,t); swap(t[i],t[x]); } } } void dfs2( ll x, ll cnt, string t ) { if( cnt > n-1 || x == t.length()-1 ) { //debug(cnt), debug(tmax), debug(strmax), debug(t); strmax = max(strmax,t); return ; } if( t[x] == tmax[x] ) { dfs2(x+1,cnt,t); return ; } char c = ‘0‘; bool flag = true; for( ll i = x+1; i < t.length(); i ++ ) { if( t[i] >= c ) { c = t[i]; flag = false; } } for( ll i = x+1; i < t.length(); i ++ ) { if( t[i] == c ) { swap(t[i],t[x]); dfs2(x+1,cnt+1,t); swap(t[i],t[x]); } } } string rev( string s ) { string t = ""; for( ll i = 0, j = s.length()-1; i < s.length(); i ++, j -- ) { t = t + s[j]; } return t; } int main() { ios::sync_with_stdio(0),cin.tie(0),cout.tie(0); ll T, t = 1; cin >> T; while( T -- ) { s = ""; ll j = 0; cin >> m >> n; while(m) { char c = (m%10)+‘0‘; s = s + c; m /= 10; } s = rev(s); tmin = s, tmax = s; sort(tmin.begin(),tmin.end()); sort(tmax.begin(),tmax.end(),cmp); if( tmin[0] == ‘0‘ ) { char c = ‘a‘; ll inx = -1; for( ll i = 1; i < tmin.length(); i ++ ) { if( tmin[i] != ‘0‘ && tmin[i] < c ) { c = tmin[i]; inx = i; } } if( inx != -1 ) { swap(tmin[inx],tmin[0]); } } if( n >= s.length()-1 ) { cout << tmin << " " << tmax << endl; } else { strmin = s; dfs1(0,0,strmin); strmax = s; dfs2(0,0,strmax); cout << strmin << " " << strmax << endl; } } return 0; } /* 123112 2 111322 322111 10001 2 */
Beautiful Now 杭電第五場 暴力枚舉