A
/* Huyyt */ #include <bits/stdc++.h> #define mem(a,b) memset(a,b,sizeof(a)) #define mkp(a,b) make_pair(a,b) #define pb push_back const int dir[8][2] = {{0, 1}, {1, 0}, {0, -1}, { -1, 0}, {1, 1}, {1, -1}, { -1, -1}, { -1, 1}}; using namespace std; typedef long long ll; int main() { ios_base::sync_with_stdio(0); cin.tie(0); ll n; cin >> n; if(n%10<=5) { n-=n%10; } else { n+=10-n%10; } cout<<n<<endl; return 0; }
B
/* Huyyt */ #include <bits/stdc++.h> #define mem(a,b) memset(a,b,sizeof(a)) #define mkp(a,b) make_pair(a,b) #define pb push_back const int dir[8][2] = {{0, 1}, {1, 0}, {0, -1}, { -1, 0}, {1, 1}, {1, -1}, { -1, -1}, { -1, 1}}; using namespace std; typedef long long ll; int main() { ios_base::sync_with_stdio(0); cin.tie(0); ll n; cin >> n; ll a,b; cin >> a >> b; ll bei=n/a; ll now; for(ll i=0;i<=bei;i++) { now=n-i*a; if(now%b==0) { cout<<"YES"<<endl; cout<<i<<" "; cout<<now/b<<endl; return 0; } } cout<<"NO"<<endl; return 0; }
C
/* Huyyt */ #include <bits/stdc++.h> #define mem(a,b) memset(a,b,sizeof(a)) #define mkp(a,b) make_pair(a,b) #define pb push_back const int dir[8][2] = {{0, 1}, {1, 0}, {0, -1}, { -1, 0}, {1, 1}, {1, -1}, { -1, -1}, { -1, 1}}; using namespace std; typedef long long ll; map<int, string> mpb; map<string, int> mp; string p[25][205]; int ok[25][205]; int num[25]; int ans[25]; int pop = 0; bool check(string a, string b) { if (b.size() > a.size()) { return false; } for (int i = 0; i < b.size(); i++) { if (a[a.size() - i - 1] != b[b.size() - i - 1]) { return false; } } return true; } int main() { ios_base::sync_with_stdio(0); cin.tie(0); int flag; int n; cin >> n; int now; string ch; string ch2; for (int i = 1; i <= n; i++) { cin >> ch >> now; if (!mp[ch]) { mp[ch] = ++pop; mpb[pop] = ch; } for (int j = 1; j <= now; j++) { cin >> ch2; flag=1; for(int k=1;k<=num[mp[ch]];k++) { if(ch2==p[mp[ch]][k]) { flag=0; break; } } if(flag) p[mp[ch]][++num[mp[ch]]] = ch2; } } for (int i = 1; i <= pop; i++) { ans[i] = num[i]; //cout<<ans[i]<<endl; for (int j = 1; j <= num[i]; j++) { for (int k = 1; k <= num[i]; k++) { if (j == k) { continue; } if (check(p[i][k], p[i][j])) { ok[i][j] = 1; ans[i]--; break; } } } } cout << pop << endl; for (int i = 1; i <= pop; i++) { cout << mpb[i] << " "; cout << ans[i] << " "; for (int j = 1; j <= num[i]; j++) { if (!ok[i][j]) { cout << p[i][j] << " "; } } cout << endl; } return 0; }
D
/*Huyyt*/ #include<bits/stdc++.h> #define mem(a,b) memset(a,b,sizeof(a)) #define pb push_back using namespace std; typedef long long ll; typedef unsigned long long ull; const ll LLmaxn = 2e18; const int N = 1e6 + 5; int num[N]; int visit[N]; int main() { ios_base::sync_with_stdio(0); cin.tie(0); int n, m, k; cin >> n >> m >> k; for (int i = 1; i <= n; i++) { cin >> num[i]; visit[num[i]] = 1; } int anser = 0; int now = 0; for (int i = 1; i <= 1000000; i++) { if (!visit[i]) { if ((i - m + 1 >= 1) && visit[i - m + 1]) { now--; } continue; } if (now < k - 1) { now++; } else { visit[i] = 0; //cout << i << endl; anser++; } if ((i - m + 1 >= 1) && visit[i - m + 1]) { now--; } } cout << anser << endl; return 0; }
E
/*Huyyt*/ #include<bits/stdc++.h> #define mem(a,b) memset(a,b,sizeof(a)) #define pb push_back using namespace std; typedef long long ll; typedef unsigned long long ull; const ll LLmaxn = 2e18; const int N = 2e5 + 5; ll num[N]; ll cha[N]; priority_queue<ll, vector<ll>, greater<ll> >q; int main() { ios_base::sync_with_stdio(0); cin.tie(0); int n; cin >> n; int sum = 0; int zero = 0; for (int i = 1; i <= n; i++) { cin >> num[i]; ll s = sqrt(num[i]); cha[i] = min(abs(s * s - num[i]), (s + 1) * (s + 1) - num[i]); if (cha[i] == 0) { sum++; } else { q.push(cha[i]); } if (num[i] == 0) { zero++; } } if (sum >= n / 2) { if (sum == n / 2) { cout << 0 << endl; return 0; } if (zero <= n / 2) { cout << sum - n / 2 << endl; } else { cout << sum - zero + (zero - n / 2) * 2 << endl; } return 0; } int t = n / 2 - sum; ll anser = 0; while (t--) { anser += q.top(); q.pop(); } cout << anser << endl; return 0; }
F
题意为给你一个长度不超过1e6的数字串 要求你插入'+'和'='使得串被分为三个无前导零的数 a,b,c 且a+b=c
解:
分情况讨论 有且只有四种情况
①/② a.size/b.size=c.size ③/④ a.size/b.size=c.size-1 枚举的复杂度为n
每次直接用暴力来check的话 很明显check的复杂度是len 而len和n是一个级别的 总复杂度就是n2 不行
所以我们需要hash一下 先用hash判断 如果符合的话 再用字符串模拟大数check
#include <iostream> #include <cstring> #include <cmath> #include <cstdio> #include <algorithm> using namespace std; const int M = 1e6 + 44; const int sed = 10; const int mod = 1e9 + 7; int n; char str[M]; //原数字串 int s[M]; //数字数组 int suf[M]; // int bas[M]； int inv[M]; long long pr(int a, int b) //快速幂 { long long r = 1, base = a; while (b != 0) { if (b & 1) { r = (r * base) % mod; } base = (base * base) % mod; b >>= 1; } return r; } bool check(int la, int lb, int lc) { int sava = la, savb = lb, savc = lc; if (la > lc || lb > lc || la < 1 || lb < 1 || lc < 1) { return false; } int lia = 1, lib = la + 1, lic = la + lb + 1; if ((la > 1 && s[lia] == 0) || (lb > 1 && s[lib] == 0) || (lc > 1 && s[lic] == 0)) { return false; } int ria = la, rib = la + lb, ric = la + lb + lc; int vala, valb, valc, up = 0, tmp; while (lc > 0) { if (la > 0) { vala = s[ria]; } else { vala = 0; } if (lb > 0) { valb = s[rib]; } else { valb = 0; } tmp = up + vala + valb; up = tmp / 10; tmp %= 10; if (tmp != s[ric]) { return false; } la--; lb--; lc--; ria--; rib--; ric--; } if (up) { return false; } for (int i = 1; i <= n; i++) { printf("%d", s[i]); if (i == sava) { printf("+"); } if (i == sava + savb) { printf("="); } } return true; } int getval(int li, int ri) { int ret, tmp; tmp = (0ll + suf[li] - suf[ri + 1]) % mod; tmp = (1ll * tmp * inv[n - ri]) % mod; ret = (tmp % mod + mod) % mod; return ret; } bool solve(int la, int lb, int lc) { int sava = la, savb = lb, savc = lc; if (la > lc || lb > lc || la < 1 || lb < 1 || lc < 1) { return false; } int lia = 1, lib = la + 1, lic = la + lb + 1; if ((la > 1 && s[lia] == 0) || (lb > 1 && s[lib] == 0) || (lc > 1 && s[lic] == 0)) { return false; } int ria = la, rib = la + lb, ric = la + lb + lc; int vala, valb, valc; vala = getval(lia, ria); valb = getval(lib, rib); valc = getval(lic, ric); // cout<<la<<' '<<lb<<' '<<lc<<endl; // cout<<vala<<' '<<valb<<' '<<valc%mod<<endl; if ((0ll + vala + valb) % mod == valc % mod && check(la, lb, lc)) { return true; } else { return false; } } int main() { scanf("%s", str + 1); n = strlen(str + 1); for (int i = 1; i <= n; i++) { s[i] = str[i] - '0'; } bas[0] = 1; for (int i = 1; i <= n; i++) //每一位的hash值 { bas[i] = 1ll * bas[i - 1] * sed % mod; } for (int i = 0; i <= n; i++) //处理逆元 { inv[i] = pr(bas[i], mod - 2); } suf[n + 1] = 0; for (int i = n; i >= 1; i--) //从后往前hash { suf[i] = (suf[i + 1] + 1ll * bas[n - i] * s[i] % mod) % mod; } for (int i = 1; i <= n; i++) //枚举 { //四种情况 如果有的话就结束 //lc-1,lc-1,lc lc,lc,lc lc if (solve(i - 1, n - i - (i - 1), i) || solve(i, n - i - i, i) || solve(n - i - (i - 1), i - 1, i) || solve(n - i - i, i, i)) { return 0; } } return 0; }