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乌拉5个牛币也是币~,欸嘿嘿。,
题目看起来是竖着输出,但是很显然,步长改为2即可输出原字符串/。
#pragma GCC optimize("O2,O3,Ofast,inline,unroll-all-loops,-ffast-math") #pragma GCC optimize("O2,O3,Ofast,inline,unroll-all-loops,-ffast-math") #include <bits/stdc++.h> using namespace std; #define js ios::sync_with_stdio(false);cin.tie(0); cout.tie(0) typedef long long ll; typedef unsigned long long ull; typedef long double ld; const ll MOD = 1e9 + 7; inline ll read() { ll s = 0, w = 1; char ch = getchar(); for (; !isdigit(ch); ch = getchar()) if (ch == '-') w = -1; for (; isdigit(ch); ch = getchar()) s = (s << 1) + (s << 3) + (ch ^ 48); return s * w; } inline void write(ll x) { if (!x) { putchar('0'); return; } char F[40]; ll tmp = x > 0 ? x : -x; if (x < 0)putchar('-'); int cnt = 0; while (tmp > 0) { F[cnt++] = tmp % 10 + '0'; tmp /= 10; } while (cnt > 0)putchar(F[--cnt]); } inline ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; } ll qpow(ll a, ll b) { ll ans = 1; while (b) { if (b & 1) ans *= a; b >>= 1; a *= a; } return ans; } ll qpow(ll a, ll b, ll mod) { ll ans = 1; while (b) { if (b & 1)(ans *= a) %= mod; b >>= 1; (a *= a) %= mod; }return ans % mod; } inline int lowbit(int x) { return x & (-x); } const int N = 1e5 + 7; string s; int main() { js; cin >> s; int len = s.size(); for (int i = 0; i < len; i += 2) cout << s[i]; for (int i = 1; i < len; i += 2) cout << s[i]; return 0; }