/tags/div2/ Recent content in Div2 on shift's Blog Hugo -- gohugo.io zh Tue, 06 Feb 2024 10:07:37 +0800 /posts/acm/educational-codeforces-round-161-rated-for-div.-2/ Tue, 06 Feb 2024 10:07:37 +0800 /posts/acm/educational-codeforces-round-161-rated-for-div.-2/ A. Tricky Template 我们对每个位置 $i$ 来看,只要 $a_i == c_i \ or \ b_i == c_i$​ ,那么就会使其不成立, 如果是整个字符串呢，那么就是，那么就需要每个位置都成立才能使其不成立，于是遍历判断一下即可 $code:$ void solve() { int n; std::cin >> n; std::string a, b, c; std::cin >> a >> b >> c; int ok = 0; for(int i = 0; i < n; i ++ ) { if(a[i] != c[i] && b[i] != c[i]) { ok = 1; } } std::cout << (ok ? "YES" : "NO")