PTA2022 520钻石争霸赛题解
7-1 520表白
不用说
#include<bits/stdc++.h> using namespace std; typedef long long ll; const int maxn = 1e9; const int maxm = 1e5 + 5; const int inf = 2147483647; using namespace std; int solve(){ int n, m; cin >> n; cout << n << "! " << "520!"; return 0; } int main() { int t = 1; // cin >> t; while(t--){ solve(); } return 0; }
7-2 分糖豆
不用说
#include<bits/stdc++.h> using namespace std; typedef long long ll; const int maxn = 1e9; const int maxm = 1e5 + 5; const int inf = 2147483647; using namespace std; int solve(){ int n, m, k; cin >> n >> m >> k; if(n * k == m){ cout << "zheng hao mei ren " << k << "!"; }else if(n * k < m){ cout << "hai sheng " << m - n * k << "!"; }else cout << "hai cha " << n * k - m << "!"; return 0; } int main() { int t = 1; // cin >> t; while(t--){ solve(); } return 0; }
7-3 约会App
判断下就可以，不用说
#include<bits/stdc++.h> using namespace std; typedef long long ll; const int maxn = 1e9; const int maxm = 1e5 + 5; const int inf = 2147483647; using namespace std; int solve(){ int x, a, b, c, d, n; cin >> x >> a >> b >> c >> d >> n; while(n--){ int s, y, h; cin >> s >> y >> h; if(s != x && y >= a && y <= b && h >= c && h <= d){ cout << s << ' ' << y << ' ' << h << "\n"; } } return 0; } int main() { int t = 1; // cin >> t; while(t--){ solve(); } return 0; }
7-4 关于奇数的等式
范围小从1开始找就行，根据等式判断
#include<bits/stdc++.h> using namespace std; typedef long long ll; const int maxn = 1e9; const int maxm = 1e5 + 5; const int inf = 2147483647; using namespace std; int solve(){ int n, m, x, y, z, f = 0; cin >> n >> m; for(int i = 1; i <= m - 4; i += 2){ for(int j = i + 2; j <= m - 2; j += 2){ for(int k = j + 2; k <= m; k += 2){ if((1.0 * 3 / n) == ((1.0 / i) + (1.0 / j) + (1.0 / k))){ f = 1; cout << i << ' ' << j << ' ' << k; return 0; } } } } if(!f)cout << "No solution in (3, " << m << "]."; return 0; } int main() { int t = 1; // cin >> t; while(t--){ solve(); } return 0; }
7-5 我侬数
#include<bits/stdc++.h> using namespace std; typedef long long ll; const int maxn = 1e9; const int maxm = 1e5 + 5; const int inf = 2147483647; using namespace std; int aa[10], cc[10];//分别是存 原本基础数 和 需要判断的数 的0-9的个数 int solve(){ string a, b, c, d; cin >> a >> b; int i = 0, j = 0, p = 0; //去掉前面0 while(a[i] == '0' && i < a.length())i++; while(b[j] == '0' && j < b.length())j++; //找0-9的个数 while(i < a.length())aa[a[i] - '0']++, i++; while(j < b.length())aa[b[j] - '0']++, j++; while(1){ for(int k = 0; k < 10; k ++)cc[k] = 0;//每次都要重新清0 cin >> c >> d; if(c == "0" && d == "0")break; //同上 i = 0, j = 0; while(c[i] == '0' && i < c.length())i++; while(d[j] == '0' && j < d.length())j++; while(i < c.length())cc[c[i] - '0']++, i++; while(j < d.length())cc[d[j] - '0']++, j++; int f = 0; for(int k = 0; k < 10; k++){ //有不一样就No if(aa[k] != cc[k]){ cout << "No\n"; f = 1; break; } } if(!f)cout << "Yes\n"; } return 0; } int main() { int t = 1; // cin >> t; while(t--){ solve(); } return 0; }
7-6 非诚勿扰
#include<bits/stdc++.h> using namespace std; typedef long long ll; const int maxn = 1e9; const int maxm = 1e5 + 5; const int inf = 2147483647; using namespace std; int solve(){ int n, m, k, j = 3, p = 3, f = 0, q, h; cin >> n; k = n / 2.718; for(int i = 1; i <= n; i ++){ cin >> m; //p是找全部人中最合适的，h保留下标 if(m >= p){ p = m; h = i; } if(i <= k) j = max(j, m); else if(!f && m >= j) {q = i;f = 1;}//找决定牵手的，q存下标 } if(!f)q = 0; cout << q << ' ' << h; return 0; } int main() { int t = 1; // cin >> t; while(t--){ solve(); } return 0; }
7-7 新式六合彩
#include<bits/stdc++.h> using namespace std; typedef long long ll; const int maxn = 1e9; const int maxm = 1e5 + 5; const int inf = 2147483647; using namespace std; int a[1010][1010], minm = inf; int solve(){ int n, m; cin >> n >> m; for(int i = 1; i <= n; i ++) for(int j = 1; j <= m; j ++) cin >> a[i][j]; int r, c, x; cin >> r >> c >> x; for(int i = 1; i <= n; i ++){ for(int j = 1; j <= m; j ++){ if(i == r || j == c){ if(a[i][j] != -1) //找r行或c列中差值最小的 minm = min(minm, abs(a[i][j] - x)); } } } //范围小，一行一行输出就行，符合输出要求 for(int i = 1; i <= n; i ++){ for(int j = 1; j <= m; j ++){ if(i == r || j == c){ if(a[i][j] != -1) if((abs(a[i][j] - x)) == minm) cout << "(" << i << ":" << j << ")" << "\n"; } } } return 0; } int main() { int t = 1; // cin >> t; while(t--){ solve(); } return 0; }
7-8没时间做
