結果

問題 No.131 マンハッタン距離
ユーザー lskjfsdhj
提出日時 2019-09-01 17:48:36
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 1,573 ms / 5,000 ms
コード長 5,487 bytes
コンパイル時間 1,308 ms
コンパイル使用メモリ 115,252 KB
実行使用メモリ 5,888 KB
最終ジャッジ日時 2024-11-28 20:10:27
合計ジャッジ時間 4,530 ms
ジャッジサーバーID
(参考情報)
judge1 / judge5
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 3
other AC * 24
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

#include<iostream>
#include <iomanip>
#include <vector>
#include <algorithm>
#include <string>
#include <string.h>
#include <climits>
#include <cmath>
#include <map>
#include <set>
#include <bitset>
#include <stack>
#include <queue>
#include <fstream>
#include <functional>
using namespace std;
using ll = long long;
// graph
const ll MAX_V = 100001;
const ll MAX_E = 2;
struct edge {
ll from, to, cost;
};
edge ES[MAX_E];
vector<edge> G[MAX_V];
ll d[MAX_V];
ll prev_path[MAX_V];
ll V, E;
const ll MOD = (ll) 1e9 + 7;
const int MAX_INT = 1 << 17;
vector<bool> prime;
#define all(x) (x).begin(),(x).end()
#define PRI(n) cout << n <<endl;
#define PRI2(n, m) cout << n << " " << m << " "<<endl;
#define REP(i, n) for(int i = 0; i < (ll)n; ++i)
#define REPbit(bit, n) for(int bit = 0; bit < (int)(1<<n); ++bit)
#define FOR(i, t, n) for(ll i = t; i <= (ll)n; ++i)
void Era(int x) {
prime.resize(x + 1, 1);
prime[0] = 0;
prime[1] = 0;
FOR(i, 2, sqrt(x)) {
if (prime[i]) {
for (int j = 2 * i; j <= x; j += i) {
prime[j] = 0;
}
}
}
}
bool isPrime(ll x) {
if (x == 0)return 0;
if (x == 1)return 0;
if (x == 2)return 1;
if (x % 2 == 0)return 0;
FOR(i, 3, sqrt(x) + 1) {
if (x % i == 0)return 0;
}
return 1;
}
ll GCD(ll a, ll b) {
if (b == 0)return a;
return GCD(b, a % b);
}
ll LCM(ll a, ll b) {
ll gcd = GCD(a, b);
return a / gcd * b;
}
ll nCr(ll n, ll r) {
vector<ll> C(r + 1);
C[0] = 1;
FOR(i, 1, n)for (ll j = min(i, r); j < 1; --j)
C[j] = (C[j] + C[j - 1]);
return C[r];
}
template<class T>
class SegTree {
int n;
vector<T> data;
T def;
function<T(T, T)> operation;
function<T(T, T)> update;
T _query(int a, int b, int k, int l, int r) {
if (r <= a || b <= l) return def;
if (a <= l && r <= b)
return data[k];
else {
T c1 = _query(a, b, 2 * k + 1, l, (l + r) / 2);
T c2 = _query(a, b, 2 * k + 2, (l + r) / 2, r);
return operation(c1, c2);
}
}
public:
SegTree(size_t _n, T _def, function<T(T, T)> _operation,
function<T(T, T)> _update)
: def(_def), operation(_operation), update(_update) {
n = 1;
while (n < _n) {
n *= 2;
}
data = vector<T>(2 * n - 1, def);
}
void change(int i, T x) {
i += n - 1;
data[i] = update(data[i], x);
while (i > 0) {
i = (i - 1) / 2;
data[i] = operation(data[i * 2 + 1], data[i * 2 + 2]);
}
}
T query(int a, int b) {
return _query(a, b, 0, 0, n);
}
T operator[](int i) {
return data[i + n - 1];
}
};
struct UnionFind {
vector<int> par;
vector<int> rank;
UnionFind(int N) {
for (int i = 0; i < N; ++i) {
par.push_back(i);
rank.push_back(0);
}
}
int find(int x) {
if (par[x] == x)return x;
else return par[x] = find(par[x]);
}
void unite(int x, int y) {
x = find(x);
y = find(y);
if (x == y)return;
if (rank[x] < rank[y])par[x] = y;
else {
par[y] = x;
if (rank[x] == rank[y])rank[x]++;
}
}
bool same(int x, int y) {
return find(x) == find(y);
}
};
struct ListNode {
int val;
ListNode *next;
ListNode(int x) : val(x), next(NULL) {}
};
void Bellman_short(int s) {
REP(i, V)d[i] = 1LL << 50;
d[s] = 0;
REP(i, V)
REP(i, E) {
edge e = ES[i];
if (d[e.from] != 1LL << 50 && d[e.to] > d[e.from] + e.cost) {
d[e.to] = d[e.from] + e.cost;
}
}
}
bool Bellman_negLoop(int s) {
REP(i, V)d[i] = 1LL << 50;
d[s] = 0;
REP(i, V)
REP(j, E) {
edge e = ES[j];
if (d[e.from] != 1LL << 50 && d[e.to] > d[e.from] + e.cost) {
d[e.to] = d[e.from] + e.cost;
if (i == V - 1)return true;
}
}
return false;
}
void dijkstra(int s) {
typedef pair<ll, ll> P;
priority_queue<P, vector<P>, greater<P>> Q;
fill(d, d + V, LLONG_MAX);
fill(prev_path, prev_path + V, -1);
d[s] = 0;
Q.push(P(0, s));
while (!Q.empty()) {
P p = Q.top();
Q.pop();
ll v = p.second;
if (d[v] < p.first)continue;
for (edge e:G[v]) {
if (d[e.to] > d[v] + e.cost) {
d[e.to] = d[v] + e.cost;
Q.push(P(d[e.to], e.to));
prev_path[e.to] = v;
}
}
}
}
vector<ll> getPath(int t) {
vector<ll> path;
for (; t != -1; t = prev_path[t]) {
path.push_back(t);
}
reverse(all(path));
return path;
}
//
//int A[100001];
//
//bool dfs(ll x, int a) {
// A[x] = a;
// for (edge e : G[x]) {
// if (e.cost % 2 == 0) {
// if (A[e.to] == -a)return false;
// if (A[e.to] == 0 && !dfs(e.to, a))return false;
// } else {
// if (A[e.to] == a)return false;
// if (A[e.to] == 0 && !dfs(e.to, -a))return false;
// }
// }
// return true;
//}
int x, y, D;
int main() {
cin >> x >> y >> D;
if (x + y < D) {
PRI(0)
return 0;
}
int ans = 0;
REP(i, D + 1) {
if (D - i <= x && i <= y)ans++;
}
PRI(ans)
return 0;
}
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