L1-062 幸运彩票
Statement
Metadata
- 作者: 陈越
- 单位: 浙江大学
- 代码长度限制: 16 KB
- 时间限制: 400 ms
- 内存限制: 64 MB
彩票的号码有 6 位数字,若一张彩票的前 3 位上的数之和等于后 3 位上的数之和,则称这张彩票是幸运的。本题就请你判断给定的彩票是不是幸运的。
输入格式
输入在第一行中给出一个正整数 N(
输出格式
对每张彩票,如果它是幸运的,就在一行中输出 You are lucky!
;否则输出 Wish you good luck.
。
输入样例
输出样例
Solution
#include <bits/stdc++.h>
using namespace std;
#define endl "\n"
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
#define mkp make_pair
#define all(x) (x).begin(), (x).end()
using db = double;
using ll = long long;
using ull = unsigned long long;
using pII = pair<int, int>;
using pLL = pair<ll, ll>;
constexpr int mod = 1e9 + 7;
template <class T1, class T2>
inline void chadd(T1 &x, T2 y, int Mod = mod) {
x += y;
while (x >= Mod) x -= Mod;
while (x < 0) x += Mod;
}
template <class T1, class T2>
inline void chmax(T1 &x, T2 y) {
if (x < y)
x = y;
}
template <class T1, class T2>
inline void chmin(T1 &x, T2 y) {
if (x > y)
x = y;
}
inline int nextInt() {
int x;
cin >> x;
return x;
}
void rd() {}
template <class T, class... Ts>
void rd(T &arg, Ts &...args) {
cin >> arg;
rd(args...);
}
#define dbg(x...) \
do { \
cout << "\033[32;1m" << #x << " -> "; \
err(x); \
} while (0)
void err() {
cout << "\033[39;0m" << endl;
}
template <class T, class... Ts>
void err(const T &arg, const Ts &...args) {
cout << arg << ' ';
err(args...);
}
template <template <typename...> class T, typename t, typename... A>
void err(const T<t> &arg, const A &...args) {
for (auto &v : arg) cout << v << ' ';
err(args...);
}
void ptt() {
cout << endl;
}
template <class T, class... Ts>
void ptt(const T &arg, const Ts &...args) {
cout << ' ' << arg;
ptt(args...);
}
template <class T, class... Ts>
void pt(const T &arg, const Ts &...args) {
cout << arg;
ptt(args...);
}
void pt() {}
template <template <typename...> class T, typename t, typename... A>
void pt(const T<t> &arg, const A &...args) {
for (int i = 0, sze = arg.size(); i < sze; ++i) cout << arg[i] << " \n"[i == sze - 1];
pt(args...);
}
inline ll qpow(ll base, ll n) {
assert(n >= 0);
ll res = 1;
while (n) {
if (n & 1)
res = res * base % mod;
base = base * base % mod;
n >>= 1;
}
return res;
}
// head
constexpr int N = 1e5 + 10;
int n;
void run() {
int x;
rd(x);
vector<int> vec(6, 0);
for (int i = 0; i < 6; ++i) {
vec[i] = x % 10;
x /= 10;
}
if (vec[0] + vec[1] + vec[2] == vec[3] + vec[4] + vec[5])
pt("You are lucky!");
else
pt("Wish you good luck.");
}
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
cout.tie(nullptr);
cout << fixed << setprecision(20);
int _T = nextInt();
while (_T--) run();
// for (int kase = 1; kase <= _T; ++kase) {
// cout << "Case #" << kase << ": ";
// run();
// }
// while (cin >> n) run();
// run();
return 0;
}
Last update: May 4, 2022