B. Valued Keys
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You found a mysterious function f. The function takes two strings s1 and s2. These strings must consist only of lowercase English letters, and must be the same length.
The output of the function f is another string of the same length. The i-th character of the output is equal to the minimum of the i-th character of s1 and the i-th character of s2.
For example, f("ab", "ba") = "aa", and f("nzwzl", "zizez") = "niwel".
You found two strings x and y of the same length and consisting of only lowercase English letters. Find any string zsuch that f(x, z) = y, or print -1 if no such string z exists.
The first line of input contains the string x.
The second line of input contains the string y.
Both x and y consist only of lowercase English letters, x and y have same length and this length is between 1 and 100.
If there is no string z such that f(x, z) = y, print -1.
Otherwise, print a string z such that f(x, z) = y. If there are multiple possible answers, print any of them. The string zshould be the same length as x and y and consist only of lowercase English letters.
abaa
ba
nzwzlniwel
xiyez
abba
-1
The first case is from the statement.
Another solution for the second case is "zizez"
There is no solution for the third case. That is, there is no z such that f("ab", z) = "ba".
解题说明:此题是一道字符串比较题,如果x[i]<y[i],那么肯定不存在,输出-1。如果所有的x[i]都>=y[i],那么y就是其中一个z,直接输出y就可以了。
#include<cstdio>#include<algorithm>#include<cstring>#include<cstdlib>#include<iostream>using namespace std;int main(){ char x[101],y[101];int i=0; scanf("%s",x); scanf("%s",y); while(x[i]) { if(x[i]<y[i]) {break;}i++;} if(x[i] == 0){printf("%s\n",y);}else {printf("-1\n");}return 0;}
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