A
#include <bits/stdc++.h> #define PI acos(-1.0) #define mem(a,b) memset((a),b,sizeof(a)) #define TS printf("!!! ") #define pb push_back #define inf 1e9 //std::ios::sync_with_stdio(false); using namespace std; //priority_queue<int,vector<int>,greater<int>> que; get min const double eps = 1.0e-8; typedef pair<int, int> pairint; typedef long long ll; typedef unsigned long long ull; const int maxn = 3e7 + 10; const int maxm = 300; const int turn[4][2] = {{1, 0}, { -1, 0}, {0, 1}, {0, -1}}; //priority_queue<int, vector<int>, less<int>> que; //next_permutation ll mod = 1e9 + 7; string ans; int n; string a; int flag; bool ok(char x) { if (x == 'a' || x == 'e' || x == 'i' || x == 'o' || x == 'u' || x == 'y') { return true; } return false; } int main() { cin >> n >> a; flag = 0; for (int i = 0; i < n; i++) { if (flag) { if (ok(a[i])) { continue; } else { ans += a[i]; flag = 0; } } else { if (ok(a[i])) { flag = 1; ans += a[i]; } else { ans += a[i]; flag = 0; } } } cout << ans << endl; return 0; }
B
#include <bits/stdc++.h> #define PI acos(-1.0) #define mem(a,b) memset((a),b,sizeof(a)) #define TS printf("!!! ") #define pb push_back #define inf 1e9 //std::ios::sync_with_stdio(false); using namespace std; //priority_queue<int,vector<int>,greater<int>> que; get min const double eps = 1.0e-8; typedef pair<int, int> pairint; typedef long long ll; typedef unsigned long long ull; const int maxn = 3e7 + 10; const int maxm = 300; const int turn[4][2] = {{1, 0}, { -1, 0}, {0, 1}, {0, -1}}; //priority_queue<int, vector<int>, less<int>> que; //next_permutation ll mod = 1e9 + 7; int number[1000005]; int anser = 0; int main() { int n; cin >> n; for (int i = 1; i <= n; i++) { int cur; cin >> cur; number[cur] = 1; } for (int i = 2; i <= 500000; i++) { if (number[i]) { anser = max(anser, i - 1); } } for (int i = 999999; i >= 500001; i--) { if (number[i]) { anser = max(anser, 1000000 - i); } } cout << anser << endl; return 0; }
C
N*N的方格里 每个方格可以填0或者1 要求每个M*M的小方格里面必须要有一个0
题目给你一个数 问你有没有一对 N,M使之成立
总共有N^2个方格 因为每个M*M的小方格里面需要有一个0 所以每行和每列最少都需要N/M个0 总共就是（N/M）*（N/M）个0
所以答案就是N^2-（N/M）^2个 利用立方差公式可以分解为 (N+(N/M)*(N-(N/M)=ANS 也就是分解质因数 复杂度T*SQRT X
每次遇到一个N M都要检测是否符合条件
#include <bits/stdc++.h> #define PI acos(-1.0) #define mem(a,b) memset((a),b,sizeof(a)) #define TS printf("!!! ") #define pb push_back #define inf 1e9 //std::ios::sync_with_stdio(false); using namespace std; //priority_queue<int,vector<int>,greater<int>> que; get min const double eps = 1.0e-10; const double EPS = 1.0e-4; typedef pair<int, int> pairint; typedef long long ll; typedef unsigned long long ull; //const int maxn = 3e5 + 10; const int maxm = 300; const int turn[4][2] = {{1, 0}, { -1, 0}, {0, 1}, {0, -1}}; //priority_queue<int, vector<int>, less<int>> que; //next_permutation int main() { int n; cin >> n; while (n--) { int flag = 0; int x; cin >> x; if (x == 0) { cout << 1 << " " << 1 << endl; continue; } for (int i = 1; i <= ((int)sqrt(x) + 1) && (!flag); i++) { if (x % i == 0) { int chu; int beichu; int a; a = x / i; chu = abs(i - a) / 2; if (chu == 0) { continue; } beichu = min(i, a) + chu; int m; m = beichu / chu; if (beichu * beichu - (beichu / m) * (beichu / m) == x) { cout << beichu << " " << m << endl; flag = 1; } //break; } } if (!flag) { cout << -1 << endl; } } }
D
给你N个点 每个点可以都办足球比赛 票价为Ai 同时有M条路（不保证图联通）每条路有COST
问你在城市i的球迷想看球需要的最少的钱（往返加球票钱）
用floyd肯定不行 需要建一个超级源点 然后上堆优化的迪杰斯特拉跑一边 超级源点与原来每个点之间连的边的COST为其票价
#include <bits/stdc++.h> #define PI acos(-1.0) #define mem(a,b) memset((a),b,sizeof(a)) #define TS printf("!!! ") #define pb push_back #define inf 1e9 //std::ios::sync_with_stdio(false); using namespace std; //priority_queue<int,vector<int>,greater<int>> que; get min const double eps = 1.0e-10; const double EPS = 1.0e-4; typedef pair<int, int> pairint; typedef long long ll; typedef unsigned long long ull; //const int maxn = 3e5 + 10; const int maxm = 300; const int turn[4][2] = {{1, 0}, { -1, 0}, {0, 1}, {0, -1}}; //priority_queue<int, vector<int>, less<int>> que; //next_permutation int n, m; priority_queue<pair<ll, int> > que; vector<pair<int, ll> > gra[200005]; ll sdis[200005]; int visit[200005]; int main() { for (int i = 1; i <= 200000; i++) { sdis[i] = 1e18; } cin >> n >> m; int from, to; ll cost; for (int i = 1; i <= m; i++) { scanf("%d %d %lld", &from, &to, &cost); cost *= 2LL; gra[from].pb(make_pair(to, cost)); gra[to].pb(make_pair(from, cost)); } for (int i = 1; i <= n; i++) { scanf("%lld", &cost); gra[0].pb(make_pair(i, cost)); } que.push({0, 0}); while (!que.empty()) { int now = que.top().second; que.pop(); if (visit[now]) { continue; } visit[now] = 1; int len = gra[now].size(); for (int i = 0; i < len; i++) { to = gra[now][i].first; ll value = gra[now][i].second; if (!visit[to] && sdis[to] > sdis[now] + value) { sdis[to] = sdis[now] + value; que.push({ -sdis[to], to}); } } } for (int i = 1; i <= n; i++) { cout << sdis[i] << " "; } cout << endl; }