A - Little C Loves 3 I
CodeForces - 1047A
题意:一个数分成三份,每一个都不是3的倍数
题解:分成 (1, 1, n - 2) 或者分成(1, 2, n - 3 )
#include <iostream> #include <cstdio> #include <cmath> #include <cstring> #include <algorithm> #include <vector> #include <queue> #include <set> #include <map> #define INF 0x3f3f3f3f #define lowbit(x) (x&(-x)) using namespace std; typedef long long ll; const int maxn = 1e3+7; int main() { ll n; scanf("%lld",&n); if((n - 2) % 3 == 0) printf("1 2 %lld\n",n - 3); else printf("1 1 %lld\n",n - 2); }
B - Cover Points
CodeForces - 1047B
题意:平面上有n个点,用一个顶点在原点,两直角边分别在x轴和y轴的等腰直角三角形覆盖这些点,问能将这些点全部覆盖的三角形的直角边最短是多长
题解:求解max(x + y)
#include <iostream> #include <cstdio> #include <cmath> #include <cstring> #include <algorithm> #include <vector> #include <queue> #include <set> #include <map> #define INF 0x3f3f3f3f #define lowbit(x) (x&(-x)) using namespace std; typedef long long ll; const int maxn = 1e3+7; int main() { int n; scanf("%d",&n); int maxx = 0; while(n--) { int a,b; scanf("%d %d",&a,&b); maxx = max(maxx,a + b); } printf("%d\n",maxx); }
C - Enlarge GCD
CodeForces - 1047C
题意:n个数的gcd是k,要你删掉最少的数使得删完后的数组的gcd > k
题解:先求出k,然后每个数除以k。然后找出出现次数最多的质因数即可。
#include <iostream> #include <cstdio> #include <cmath> #include <cstring> #include <algorithm> #include <vector> #include <queue> #include <set> #include <map> #define INF 0x3f3f3f3f #define lowbit(x) (x&(-x)) using namespace std; typedef long long ll; const int maxn = 3e5 + 10; const int M = 1.5e7 + 10; int pn; int gcd(int a,int b) { return b == 0 ? a : gcd(b,a % b); } int a[maxn]; int num[M]; int p[4000],prime[4000]; void init() { pn = 0; memset(p,0,sizeof p); for(int i=2;i<4000;i++) { if(!p[i]) prime[pn++] = i; for(ll j =0 ;j < pn && i * prime[j] < 4000; j++){ p[i * prime[j]] = 1; if(i % prime[j] == 0) continue; } } //cout<<pn<<endl; } int main() { int n,mingcd; init(); scanf("%d",&n); memset(num,0,sizeof num); for(int i=1;i<=n;i++) { scanf("%d",&a[i]); if(i == 1) mingcd = a[i]; else mingcd = gcd(mingcd,a[i]); } for(int i=1;i<=n;i++) a[i] /= mingcd; int ans = -1; for(int i=1;i<=n;i++) { for(int j=0;j<pn && prime[j] * prime[j] <= a[i];j++) { if(a[i] % prime[j] == 0) { num[prime[j]]++; ans = max(ans,num[prime[j]]); while(a[i] % prime[j] == 0) a[i] /= prime[j]; } } if(a[i] > 1) { num[a[i]]++; ans = max(ans,num[a[i]]); } } printf("%d\n", ans == -1 ? ans : n - ans); }
D - Little C Loves 3 II
CodeForces - 1047D
题意:给你n*m得棋盘,让你找两点之间距离为3的点的个数,不能重复使用,距离定义,两坐标差绝对值之和、
题解:找规律,找特殊样例即可
#include <iostream> #include <cstdio> #include <cmath> #include <cstring> #include <algorithm> #include <vector> #include <queue> #include <set> #include <map> #define INF 0x3f3f3f3f #define lowbit(x) (x&(-x)) using namespace std; typedef long long ll; const int maxn = 1e3+7; int gcd(int a,int b) { return b == 0 ? a : gcd(b,a % b); } int main() { ll n,m; ll ans; scanf("%lld %lld",&n,&m); if(n > m) swap(n,m); if(n == 1) { if(m % 6 == 0) ans = n * m; else if(m % 6 <= 3) ans = m - m % 6; else ans = m - (6 - m % 6); } else if(n == 2) { if(m == 2) ans = 0; else if(m == 3) ans = 4; else if(m == 7) ans = 12; else ans = n * m; } else { if(n * m % 2 == 1) ans = n * m - 1; else ans = n * m; } printf("%lld\n",ans); }
E - Region Separation
CodeForces - 1047E
题意:给定一棵大小为n的树,点有点权,在一个方案中可以将整棵树划分多次,要求每次划分后各个联通块的权值和相等,问有多少种划分方案
原文地址:https://www.cnblogs.com/smallhester/p/11155896.html
时间: 2024-11-14 12:52:34