#include<iostream> #include<cstdio> #include<algorithm> #include<cstring> #include<cmath> using namespace std; typedef long long ll; const int N=1e6+10,INF=1e8; int a[N]; int main() { int t; scanf("%d", &t); while(t--) { ll n,s; scanf("%lld%lld", &n,&s); n *= n; printf("%lld\n", s / n); } return 0; }
#include<iostream> #include<cstdio> #include<algorithm> #include<cstring> #include<cmath> using namespace std; typedef long long ll; const int N=2e5+10,INF=1e8; ll a[N]; int main()//红名大佬的题解 { int t; scanf("%d", &t); while(t--) { int n,f=0; scanf("%d", &n); for (int i = 1; i <= n;i++) { scanf("%lld", &a[i]); } sort(a + 1, a + 1 + n); ll l = 2, r = n, ans = a[n] - a[1] - a[2]; while(l<r) { if(ans>0) { f = 1; break; } ans += a[--r] - a[++l]; } if(f) printf("YES\n"); else printf("NO\n"); } return 0; } /*我的题解 #include<iostream> #include<cstdio> #include<algorithm> #include<cstring> #include<cmath> using namespace std; typedef long long ll; const int N=2e5+10,INF=1e8; ll a[N],s[N]; int main() { int t; scanf("%d", &t); while(t--) { int n; ll sum = 0; scanf("%d", &n); for (int i = 1; i <= n;i++) s[i] = 0; for (int i = 1; i <= n ;i++) { scanf("%lld", &a[i]); } sort(a + 1, a + 1 + n); for (int i = 1; i <= n;i++) { s[i] = s[i - 1] + a[i]; } int k = n / 2; if(n%2==0) { int i = n - 1,f=0; for (int j = 2; j <= k + 1; j++) { if(s[n]-s[i]<=s[j]) { if(i>k+1) i--; else { printf("NO\n"); break; } } else { printf("YES\n"); break; } } } else { if(s[n]-s[k+1]>s[k+1]) printf("YES\n"); else printf("NO\n"); } } return 0; } */
参考知乎,位运算的使用,对于每个数可以枚举到2的15次方,然后用n=1!+2!+3!...+t表示,最后把t拆成2的几次幂,即算一下t里有几个1加一起取最小值。
#include<iostream> #include<cstdio> #include<algorithm> #include<cstring> #include<cmath> #include<bitset> using namespace std; typedef long long ll; const int N=1e5+10,INF=1e8; ll fac[20]; int main() { int t; ll res = 1; scanf("%d", &t); for (int i = 1; i <= 15;i++) { res *= i; fac[i - 1] = res; } while (t--) { ll n; int ans = 1000; scanf("%lld", &n); for (int i = 0; i < 1 << 15; i++) { bitset<15> b(i); ll t = n; for (int j = 0; j < 15;j++) t -= fac[j] * b[j]; if(t>=0) { bitset<64> b1(t); ans = min(ans,(int)(b1.count() + b.count())); } } printf("%d\n", ans); } return 0; }