2017 Multi-University Training Contest - Team 2 1001 1 Is Derek lying?
2017-07-27 19:15
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题目链接
题意:
先给出三个数,分别表示题目的个数,Derek说自己题目正确的个数和Alfia说自己题目正确的个数,然后给出两个字符串,分别表示两个人的答案,判断是否有人说谎。
思路:
先找出两个字符串相同的个数和不同的个数,然后用一个循环将X的值和Y的值还有相同的值同时进行减,直到其中有一个值为零,这里假设对的题目全是相等的,然后进行判断,如果两个同时为零,则必定没有说谎,
否则将X与Y的值相加若小于不同的,则没说谎,否则说谎。
代码:
题意:
先给出三个数,分别表示题目的个数,Derek说自己题目正确的个数和Alfia说自己题目正确的个数,然后给出两个字符串,分别表示两个人的答案,判断是否有人说谎。
思路:
先找出两个字符串相同的个数和不同的个数,然后用一个循环将X的值和Y的值还有相同的值同时进行减,直到其中有一个值为零,这里假设对的题目全是相等的,然后进行判断,如果两个同时为零,则必定没有说谎,
否则将X与Y的值相加若小于不同的,则没说谎,否则说谎。
代码:
# include <bits/stdc++.h> using namespace std; const int MAXN = 80000+10; int main () { int T; int N, X, Y; char A[MAXN], B[MAXN]; scanf ("%d", &T); while (T--) { scanf ("%d%d%d", &N, &X, &Y); int sum1 = 0, sum2 = 0; scanf ("%s", &A); scanf ("%s", &B); for (int i=0; i<N; ++i) { if (A[i] != B[i]) sum1 ++; else sum2++; } while (X>0 && Y>0 && sum2>0) { X--; Y--; sum2--; } if (X == 0 && Y == 0) printf ("Not lying\n"); else { if (X + Y <= sum1) printf ("Not lying\n"); else printf ("Lying\n"); } } return 0; }
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