CodeForces - 584C Marina and Vasya （模拟）找规律

CodeForces - 584C Marina and Vasya Time Limit: 1000MS   Memory Limit: 262144KB   64bit IO Format: %I64d & %I64u Submit Status Description Marina loves strings of the same length and Vasya loves when there is a third string, different from them in exactly tcharacters. Help Vasya find at least one such string. More formally, you are given two strings s1, s2 of length n and number t. Let's denote as f(a, b) the number of characters in which strings a and b are different. Then your task will be to find any string s3 of length n, such that f(s1, s3) = f(s2, s3) = t. If there is no such string, print  - 1. Input The first line contains two integers n and t (1 ≤ n ≤ 105, 0 ≤ t ≤ n). The second line contains string s1 of length n, consisting of lowercase English letters. The third line contain string s2 of length n, consisting of lowercase English letters. Output Print a string of length n, differing from string s1 and from s2 in exactly t characters. Your string should consist only from lowercase English letters. If such string doesn't exist, print -1. Sample Input Input 3 2 abc xyc Output ayd Input 1 0 c b Output -1 Source Codeforces Round #324 (Div. 2) //题意： 给出两个长度为n的字符串，先要求构造出一个长度也为n的串使其满足其与这两个串在相同位置不相同的字符数有m个，如果不存在这样的串则输出-1 //思路：先找出两个串相同的字母的个数k，那么不同的个数为mm=n-k; 通过模拟可以知道当2*m<mm时肯定输出-1； 当符合情况时可以分两种： 1、当m+k>=n时：先将两串相同位置的不同的的mm个字符变成不等于它们的字符，然后肯定要有（m-mm）个位置相同的相同字符变成与它们不同的字符，剩下的都是相同的就行了。 2、当mm<2*t时：那么相同位置的相同字符直接输出即可，相同位置的不同字符有（mm-m）个a串的字符，有（mm-m）个b串字符，其余的是与a,b串都不相同的字符。（不太理解的可以自己找例子模拟一下就懂了） Hait：要注意新串的字符都是  ‘a’<=s<='z'的。（在这块WA了5次）。 #include<stdio.h> #include<string.h> #include<math.h> #include<map> #include<queue> #include<stack> #include<algorithm> #include<iostream> #define INF 0x3f3f3f3f #define ull unsigned long long #define long long #define IN __in64 #define N 100010 #define M 1000000007 using namespace std; char s[N]; char a[N],b[N]; int main() { int n,m,i,j,k,kk,mm,mk,la,lb,l; while(scanf("%d%d",&n,&m)!=EOF) { memset(s,'\0',sizeof(s)); memset(a,'\0',sizeof(a)); memset(b,'\0',sizeof(b)); scanf("%s%s",a,b); k=0; for(i=0;i<n;i++) { if(a[i]==b[i]) k++; } mm=n-k; mk=mm/2; if(2*m<mm) printf("-1\n"); else if(m+k>=n) { kk=m-mm; for(i=0;i<n;i++) { char d; if(a[i]==b[i]) { if(kk) { for(d='a';d<='z';d++) { if(d!=a[i]) { s[i]=d; break; } } kk--; } else s[i]=a[i]; } else { for(d='a';d<='z';d++) { if(d!=a[i]&&d!=b[i]) { s[i]=d; break; } } } } puts(s); } else { la=lb=mm-m;l=mm-la-lb; for(i=0;i<n;i++) { if(a[i]==b[i]) s[i]=a[i]; else { if(la) { s[i]=a[i]; la--; } else if(lb) { s[i]=b[i]; lb--; } else if(l) { char d; for(d='a';d<='z';d++) { if(d!=a[i]&&d!=b[i]) { s[i]=d; break; } } l--; } } } puts(s); } } return 0; }