結果
問題 | No.2323 Nafmo、A+Bをする |
ユーザー |
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提出日時 | 2023-05-28 13:34:39 |
言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 2 ms / 2,000 ms |
コード長 | 3,734 bytes |
コンパイル時間 | 960 ms |
コンパイル使用メモリ | 115,104 KB |
最終ジャッジ日時 | 2025-02-13 09:53:12 |
ジャッジサーバーID (参考情報) |
judge2 / judge2 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 17 |
ソースコード
#define _CRT_SECURE_NO_WARNINGS#include<iostream>#include<cstdio>#include<string>#include<cmath>#include<algorithm>#include<map>#include<vector>#include<stack>#include<iomanip>#include<queue>#include<set>#include<functional>#include<tuple>#include<bitset>#include<cassert>#include<cstdint>#include<complex>using namespace std;bool printb(bool f) {if (f)printf("Yes\n");else printf("No\n");return f;}template<class T>void prt(T t, string sep = "\n") { cout << t << sep; return; }template<class T>void printl(vector<T> a, string sep = " ") {for (int i = 0; i < a.size(); i++) {cout << a[i] << sep;}cout << "\n";return;}#define all(a) a.begin(),a.end()#define rep(i, n) for (int i = 0; i < (int)(n); i++)using llong = long long;using pii = pair<int, int>;using pll = pair<llong, llong>;using pli = pair<llong, int>;using pil = pair<int, llong>;template<typename T> using vec2 = vector<vector<T>>;template<typename T> inline bool chmin(T& a, T b) { return (a > b) ? (a = b, true) : false; }template<typename T> inline bool chmax(T& a, T b) { return (a < b) ? (a = b, true) : false; }bool bitIn(llong a, int b) { return ((a >> b) & 1); }int bitCnt(llong a) {int re = 0;while (a > 0) {if (a & 1)re++;a >>= 1;}return re;}llong powL(llong n, llong i) {llong re = 1;while (i >= 1) {if (i & 1) re *= n;n *= n;i >>= 1;}return re;}llong powL_M(llong n, llong i, llong m) {llong re = 1;while (i >= 1) {if (i & 1) {re *= n;re %= m;}n *= n;n %= m;i >>= 1;}return re;}struct point {llong x = 0, y = 0;};//lからrまでの和を返すtemplate<typename T>T sum_num(T l, T r) {if (((l + r) & 1) == 0) {return (l + r) / 2 * (l - r + 1);}else {return (l - r + 1) / 2 * (l + r);}}int dx[4] = { 0,1,0,-1 }, dy[4] = { 1,0,-1,0 };struct edge {int to, co;};static const double pi = 3.14159265358979;/*modintクラス。四則演算と累乗が定義されている。割り算はmodが素数でない時にも使える。(逆元の存在条件注意)extGCD(),GCD()を含む*//*template <class T>T extGCD(T a, T b, T& x, T& y) {if (b == 0) {x = 1;y = 0;return a;}T gcd = extGCD(b, a % b, y, x);y -= a / b * x;return gcd;}template <class T>T GCD(T a, T b) {T x, y;return extGCD(a, b, x, y);}static const int mod = 1e9 + 7; //問題文に合わせて変更することclass modint {public:long long x;modint(long long x = 0) :x((x% mod + mod) % mod) {}modint operator-() const {return (-x);}modint& operator+=(const modint& a) {if ((x += a.x) >= mod)x -= mod;return *this;}modint& operator-=(const modint& a) {if ((x += mod - a.x) >= mod) x -= mod;return *this;}modint& operator*=(const modint& a) {(x *= a.x) %= mod;return *this;}modint operator+(const modint& a) const {modint res(*this);return res += a;}modint operator-(const modint& a) const {modint res(*this);return res -= a;}modint operator*(const modint& a) const {modint res(*this);return res *= a;}modint inv() const {long long y, c;extGCD(x, (long long)mod, y, c);return y;}modint& operator/=(const modint& a) {return (*this) *= a.inv();}modint operator/ (const modint& a) const {modint res(*this);return res /= a;}friend ostream& operator<<(ostream& os, const modint& a) {os << a.x;return os;}};//pow(a , n) modint型aのn乗のmodを求めるtemplate<class T>modint powM(modint a, T n) {modint re(1);while (n > 0) {if (n & 1)re *= a;a *= a;n >>= 1;}return re;}//*/int main() {string aa, bb;cin >> aa >> bb;bitset<30> a(aa), b(bb);prt((a ^ b).to_ulong());}