結果
問題 | No.131 マンハッタン距離 |
ユーザー |
|
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 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 24 |
ソースコード
#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;// graphconst 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;}