UVA 1610 Party Games

https://vjudge.net/problem/UVA-1610

题目

给偶数个互不相同的仅包含大写字母的字符串，要求找出最短的仅包含大写字母的字符串，使一半的字符串小于等于它，一半的字符串大于它。

字符串A小于字符串B有以下情况：

1.$K=min{len(A),len(B)}, exists 0leqslant i < K, forall 0<=x<i, A[i]<B[i] and a[x]=b[x]$

2.$K=min{len(A),len(B)}, forall 0<=x<K, a[x]=b[x] and len(A)<len(B)$

其中大写字母从小到大顺序为“ABCDEFGHIJKLMNOPQRSTUVWXYZ”= =

题解

又是斗智题……

改了很多次……策略很琐碎，不想解释了……

这道题应该先写暴力搜索，再对拍，这样写起来才快！

字符串长度不确定，因此应该使用string。

AC代码

#include<bits/stdc++.h>
using namespace std;
#define REP(r,x,y) for(register int r=(x); r<(y); r++)
#define REPE(r,x,y) for(register int r=(x); r<=(y); r++)
#ifdef sahdsg
#define DBG(...) printf(__VA_ARGS__)
#else
#define DBG(...)
#endif

#define MAXN 1007
string k[MAXN];

int main() {
	#ifdef sahdsg
//	freopen("in.txt", "r", stdin);
	#endif
	int n;
	while(cin>>n && n) {
		REP(i,0,n) {
			cin >> k[i];
		}
		sort(k,k+n);
		int f=(n-1)/2;
		//f f+1
		int l1 = k[f].length(), l2 = k[f+1].length();
		if(l1>l2) {
			string ans; ans=k[f];
			REP(i,0,l2) {
				if(k[f][i]+1<k[f+1][i]) {
					ans[i]=k[f][i]+1;
					ans[i+1]=0;
					break;
				} else if(k[f][i]+1==k[f+1][i]) {
					if(i==l2-1)
						for(i++; i<l1-1; i++) {
							if(ans[i]<'Z') {
								ans[i]=k[f][i]+1;
								ans[i+1]=0;
								break;
							}
						}
					else {
						ans[i]=k[f+1][i];
						ans[i+1]=0;
					}
					break;
				}
			}
			puts(ans.c_str());
		} else { //l1<=l2 s1<s2
			string ans; ans=k[f];
			int l=k[f].length();
			REP(i,0,l1-1) {
				if(k[f][i]<k[f+1][i]) {
					ans[i]=k[f][i]+1;
					ans[i+1]=0;
					break;
				} 
			}
			puts(ans.c_str());
		}
	}
	return 0;
}

 对拍代码

#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
#include<bits/stdc++.h>
using namespace std;
FILE *f;
int main() {
	srand(time(NULL) ^ (unsigned long long)new char());
	while(1) {
		f= fopen("in.txt", "w");
		fputc('2', f); fputc('
', f);
		int k=rand()%100;
		for(int _=0; _<k; _++) {
			fputc('A'+rand()%26, f);
		}
		 fputc('
', f);
		 k=rand()%100;
		for(int _=0; _<k; _++) {
			fputc('A'+rand()%26, f);
		} fputc('
', f);
		fclose(f);
		putchar('.');
		system("oj < in.txt > out.txt");
		putchar('.');
		system("bl < in.txt > out2.txt");
		printf("[v]");putchar('
');
		if (system("fc out.txt out2.txt")) {
			putchar('!');
			putchar('a');
			getchar();
		}
	}
	return 0;
}