标签：三角形 大数 lap 输出 tip pll bitset hide ini
A.1:41:18 solved by hl
大致就是一个图上染色判断是否有矛盾。
按照样例来看，似乎存在孤立点就要输出NO
#include <map> #include <set> #include <ctime> #include <cmath> #include <queue> #include <stack> #include <vector> #include <string> #include <bitset> #include <cstdio> #include <cstdlib> #include <cstring> #include <sstream> #include <iostream> #include <algorithm> #include <functional> using namespace std; #define For(i, x, y) for(int i=x;i<=y;i++) #define _For(i, x, y) for(int i=x;i>=y;i--) #define Mem(f, x) memset(f,x,sizeof(f)) #define Sca(x) scanf("%d", &x) #define Sca2(x,y) scanf("%d%d",&x,&y) #define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z) #define Scl(x) scanf("%lld",&x) #define Pri(x) printf("%d\n", x) #define Prl(x) printf("%lld\n",x) #define CLR(u) for(int i=0;i<=N;i++)u[i].clear(); #define LL long long #define ULL unsigned long long #define mp make_pair #define PII pair<int,int> #define PIL pair<int,long long> #define PLL pair<long long,long long> #define pb push_back #define fi first #define se second typedef vector<int> VI; int read(){int x = 0,f = 1;char c = getchar();while (c<‘0‘ || c>‘9‘){if (c == ‘-‘) f = -1;c = getchar();} while (c >= ‘0‘&&c <= ‘9‘){x = x * 10 + c - ‘0‘;c = getchar();}return x*f;} const double PI = acos(-1.0); const double eps = 1e-9; const int maxn = 1010; const int maxm = 10010; const int INF = 0x3f3f3f3f; const int mod = 1e9 + 7; int N,M,X,Y; int fa[maxn],sz[maxn]; int color[maxn]; int head[maxn],tot; struct Edge{ int to,next; }edge[maxm * 2]; void init(){ for(int i = 0 ; i <= N ; i ++) fa[i] = i,sz[i] = 1; for(int i = 0; i <= N ; i ++) head[i] = -1; tot = 0; } int find(int x){ if(x == fa[x]) return x; return fa[x] = find(fa[x]); } void Union(int a,int b){ a = find(a); b = find(b); if(a != b){ sz[b] += sz[a]; fa[a] = b; } } void add(int u,int v){ edge[tot].to = v; edge[tot].next = head[u]; head[u] = tot++; } bool flag; void dfs(int x){ //cout << x << " " << color[x] << endl; for(int i = head[x]; ~i ;i = edge[i].next){ int v = edge[i].to; if(!color[v]){ color[v] = -color[x]; dfs(v); } else if(color[v] == color[x]) flag = 0; } } int main(){ while(~Sca2(N,M)){ Sca2(X,Y); init(); for(int i = 1; i <= N ; i ++) color[i] = 0; for(int i = 1; i <= M ; i ++){ int u,v ; Sca2(u,v); add(u,v); add(v,u); Union(u,v); } flag = 1; for(int i = 1; i <= X ; i ++){ int x = read(); if(color[x] == -1) flag = 0; if(!color[x]){ color[x] = 1; dfs(x); } } for(int i = 1; i <= Y; i ++){ int x = read(); if(color[x] == 1) flag = 0; if(!color[x]){ color[x] = -1; dfs(x); } } for(int i = 1; i <= N ; i ++){ // cout <<sz[find(i)] << endl; if(!color[i]){ if(sz[find(i)] == 1) flag = 0; // if(flag) puts("YES"); // else puts("NO"); color[i] = 1; dfs(i); } } if(flag) puts("YES"); else puts("NO"); } return 0; }
C.4:34:31(-4) solved by zcz
方法1.如果较大的数和较小的数比例是黄金比就是0
方法2.设b为较大的数，a为较小的数，l为b - a,满足 b > b * l - a * a >= 0就为0
要用JAVA上大数
import java.math.*; import java.util.*; import java.io.*; public class Main{ public static void main(String[] args){ Scanner in = new Scanner(System.in); BigInteger a,b,x,y,m,n,l,cnt,fu,q,zero; while(in.hasNext()){ x = in.nextBigInteger(); y = in.nextBigInteger(); a = x; b = x; b = b.max(y); a = a.min(y); l = b.subtract(a); n = b.multiply(l); q = a.multiply(a); int t = b.compareTo(n.subtract(q)); int tt = n.compareTo(q); if(t == -1 || tt == -1 || t == 0) System.out.println("1"); else System.out.println("0"); } } }
D. 1:14:59 solved by gbs
可以推公式，直接解方程解出来
x1= (X + sqrt(XX - 4ygcd(x,y) ) ) /2;
y1 = X - x1;
#include <iostream> #include<stack> #include<math.h> #include<stdlib.h> #include<string.h> #include<string> #include<ctime> #include<complex> #include<stdio.h> #include<algorithm> #include<map> #include<queue> #include<deque> using namespace std; typedef long long LL; int gcd(int a,int b) { if (b == 0) return a; return gcd(b,a%b); } int judge1(int X,int Y) { int left1 = 1; int right1 = X/2; while(left1<=right1) { //cout<<left1 int mid = (left1+right1)>>1; LL ans = 1LL*(X-mid)*mid; //cout<<ans<<‘ ‘<<Y<<‘ ‘<<mid<<endl; if (ans>Y) right1 = mid-1; else if (ans == Y) { if (gcd(mid,X-mid)!=1) return -1; return mid; } else left1 = mid+1; } return -1; } int main() { int X,Y; int gcd1; while(scanf("%d%d",&X,&Y)!=EOF){ gcd1 = gcd(X,Y); X/=gcd1; Y/=gcd1; //cout<<X<<‘ ‘<<Y<<endl; int ans1= judge1(X,Y); if (ans1 == -1) printf("No Solution\n"); else printf("%d %d\n",ans1*gcd1,(X-ans1)*gcd1); } return 0; } /* 111 5111111 11111116 12133215 251111111 111 5 15 5 25 5 14 5 25 5 13 5 25 5 12 5 25 3 14 14 5 5 14 14 5 5 */
F. 1:08:17(-1) solved by zcz and hl
WA:读错题
很显然，段数分的越多越好,同时要满足每段不同，所以从2开始分直到不能分
例如10分为 2 + 3 + 4,最后剩下的1平均从后往前分，变为 2 + 3 + 5
12分为 2 + 3 + 4 ,剩下的3分给他们变为 3 + 4 + 5
13分为 2 + 3 + 4,剩下的分为 3 + 4 + 5,又剩下1分到5上变为 3 + 4 + 6
这是一个等差数列，知道了分法就总结一下计算公式就可以了
#include <map> #include <set> #include <ctime> #include <cmath> #include <queue> #include <stack> #include <vector> #include <string> #include <bitset> #include <cstdio> #include <cstdlib> #include <cstring> #include <sstream> #include <iostream> #include <algorithm> #include <functional> using namespace std; #define For(i, x, y) for(int i=x;i<=y;i++) #define _For(i, x, y) for(int i=x;i>=y;i--) #define Mem(f, x) memset(f,x,sizeof(f)) #define Sca(x) scanf("%d", &x) #define Sca2(x,y) scanf("%d%d",&x,&y) #define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z) #define Scl(x) scanf("%lld",&x) #define Pri(x) printf("%d\n", x) #define Prl(x) printf("%lld\n",x) #define CLR(u) for(int i=0;i<=N;i++)u[i].clear(); #define LL long long #define ULL unsigned long long #define mp make_pair #define PII pair<int,int> #define PIL pair<int,long long> #define PLL pair<long long,long long> #define pb push_back #define fi first #define se second typedef vector<int> VI; int read(){int x = 0,f = 1;char c = getchar();while (c<‘0‘ || c>‘9‘){if (c == ‘-‘) f = -1;c = getchar();} while (c >= ‘0‘&&c <= ‘9‘){x = x * 10 + c - ‘0‘;c = getchar();}return x*f;} const double PI = acos(-1.0); const double eps = 1e-9; const int maxn = 1e5 + 10; const int INF = 0x3f3f3f3f; const int mod = 1e9 + 7; int N,M,K; LL fac[maxn]; LL q_p(LL a,LL b){ LL ans = 1; while(b){ if(b & 1) ans = ans * a %mod; b >>= 1; a = a * a % mod; } return ans; } LL inv(LL x){ return q_p(x,mod - 2); } void init(){ fac[0] = 1; for(int i = 1; i < maxn; i ++) fac[i] = fac[i - 1] * i % mod; } LL check(LL x){ return (x + 2) * (x - 1) / 2; } LL solve(){ LL l = 2,r = INF; LL ans = INF; while(l <= r){ LL m = l + r >> 1; if(N >= check(m)){ l = m + 1; ans = m; }else{ r = m - 1; } } return ans; } int main(){ int T; Sca(T); init(); LL inv2 = inv(2); while(T--){ Sca(N); if(N <= 4){ Pri(N); continue; } LL n = solve(); LL a = N - check(n); // cout << a << " " << n <<endl; LL ans = fac[n + 1] * inv(n + 1 - a) % mod; if(n == a){ ans = (n + 2) * fac[n] % mod *inv2 % mod; } Prl(ans); } return 0; }
G.3:05:25(-2) solved by hl
WA:没整清楚就交 + 忘记特判
点分治 + 状压dp
点分治每次寻找子树的重心，对于不同的两棵树之间维护dp1[i]表示已经被记入的树在i满足存在状态i结点的数量
dp2表示正在处理的当前子树的状态数，dp3[i]表示当前处理的子树下状态确实为i的数量
#include <map> #include <set> #include <ctime> #include <cmath> #include <queue> #include <stack> #include <vector> #include <string> #include <bitset> #include <cstdio> #include <cstdlib> #include <cstring> #include <sstream> #include <iostream> #include <algorithm> #include <functional> using namespace std; #define For(i, x, y) for(int i=x;i<=y;i++) #define _For(i, x, y) for(int i=x;i>=y;i--) #define Mem(f, x) memset(f,x,sizeof(f)) #define Sca(x) scanf("%d", &x) #define Sca2(x,y) scanf("%d%d",&x,&y) #define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z) #define Scl(x) scanf("%lld",&x) #define Pri(x) printf("%d\n", x) #define Prl(x) printf("%lld\n",x) #define CLR(u) for(int i=0;i<=N;i++)u[i].clear(); #define LL long long #define ULL unsigned long long #define mp make_pair #define PII pair<int,int> #define PIL pair<int,long long> #define PLL pair<long long,long long> #define pb push_back #define fi first #define se second typedef vector<int> VI; int read(){int x = 0,f = 1;char c = getchar();while (c<‘0‘ || c>‘9‘){if (c == ‘-‘) f = -1;c = getchar();} while (c >= ‘0‘&&c <= ‘9‘){x = x * 10 + c - ‘0‘;c = getchar();}return x*f;} const double PI = acos(-1.0); const double eps = 1e-9; const int maxn = 5e4 + 10; const int INF = 0x3f3f3f3f; const int mod = 1e9 + 7; LL N,M,K; int type[maxn]; struct Edge{ int to,next; }edge[maxn * 2]; int head[maxn],tot; void init(){ for(int i = 0 ; i <= N ; i ++) head[i] = -1; tot = 0; } void add(int u,int v){ edge[tot].to = v; edge[tot].next = head[u]; head[u] = tot++; } int root,max_part,SUM,ed; int size[maxn],vis[maxn]; void dfs_root(int u,int la){ size[u] = 1; int heavy = 0; for(int i = head[u];~i ; i = edge[i].next){ int v = edge[i].to; if(v == la || vis[v]) continue; dfs_root(v,u); heavy = max(heavy,size[v]); size[u] += size[v]; } if(max_part > max(heavy,SUM - heavy)){ max_part = max(heavy,SUM - heavy); root = u; } } LL dp1[3000],dp2[3000],dp3[3000]; void dfs_dis(int t,int la,int state){ state |= (1 << type[t]); dp3[state]++; for(int sta = state;sta; sta = (sta - 1) & state) dp2[sta]++; dp2[0]++; for(int i = head[t]; ~i ; i = edge[i].next){ int v = edge[i].to; if(vis[v] || v == la) continue; dfs_dis(v,t,state); } } LL ans; void work(int t){ int state = 1 << type[t]; for(int i = 0 ; i <= ed; i ++) dp1[i] = dp2[i] = dp3[i] = 0; for(int sta = state;sta; sta = (sta - 1) & state) dp1[sta]++; dp1[0]++; LL sum = 0; for(int i = head[t]; ~i ; i = edge[i].next){ int v = edge[i].to; if(vis[v]) continue; dfs_dis(v,t,state); for(int sta = 0; sta <= ed; sta++){ ans += dp1[sta] * dp3[ed - sta]; // cout << " " << sta <<" " << dp1[sta] << " " << dp3[ed - sta] << endl; } //ans -= sum; // sum += dp2[ed - state]; for(int sta = 0; sta <= ed; sta++){ dp1[sta] += dp2[sta]; dp2[sta] = 0; dp3[sta] = 0; } // cout << t << " " << v << " " << ans << endl; } } void divide(int t){ max_part = INF,root = t; dfs_root(t,-1); vis[root] = 1; work(root); for(int i = head[root]; ~i ; i = edge[i].next){ int v = edge[i].to; if(vis[v]) continue; SUM = size[v]; divide(v); } } int main(){ while(~scanf("%lld%d",&N,&K)){ init(); ed = (1 << K) - 1; ans = 0; for(int i = 0 ; i <= N ; i ++) vis[i] = 0; for(int i = 1; i <= N ; i ++){ Sca(type[i]); type[i]--; } for(int i = 1; i <= N - 1; i ++){ int u,v; Sca2(u,v); add(u,v); add(v,u); } if(K == 1){ Prl(N * N); continue; } SUM = N; divide(1); Prl(ans *2); } return 0; } /* 3 2 1 2 2 1 2 1 3 0 0 1 1 1 0 2 0 0 3 0 0 1 3 0 0 0 1 1 2 0 2 1 0 3 1 0 1 2 0 0 */
H.0:27:19 solved by gbs
判断一下奇偶
#include <iostream> #include<stack> #include<math.h> #include<stdlib.h> #include<string.h> #include<string> #include<ctime> #include<complex> #include<stdio.h> #include<algorithm> #include<map> #include<queue> #include<deque> using namespace std; typedef long long LL; int main() { int n; while(scanf("%d",&n)!=EOF){ if (n&1) printf("0\n"); else printf("1\n"); } return 0; } /* 111 5111111 11111116 12133215 251111111 111 5 15 5 25 5 14 5 25 5 13 5 25 5 12 5 25 3 14 14 5 5 14 14 5 5 */
I.0:16:52 solved by hl
把凸包变成一圈三角形拼起来的，
三角形面积公式是 S = 0.5absin(θ)
#include <map> #include <set> #include <ctime> #include <cmath> #include <queue> #include <stack> #include <vector> #include <string> #include <bitset> #include <cstdio> #include <cstdlib> #include <cstring> #include <sstream> #include <iostream> #include <algorithm> #include <functional> using namespace std; #define For(i, x, y) for(int i=x;i<=y;i++) #define _For(i, x, y) for(int i=x;i>=y;i--) #define Mem(f, x) memset(f,x,sizeof(f)) #define Sca(x) scanf("%d", &x) #define Sca2(x,y) scanf("%d%d",&x,&y) #define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z) #define Scl(x) scanf("%lld",&x) #define Pri(x) printf("%d\n", x) #define Prl(x) printf("%lld\n",x) #define CLR(u) for(int i=0;i<=N;i++)u[i].clear(); #define LL long long #define ULL unsigned long long #define mp make_pair #define PII pair<int,int> #define PIL pair<int,long long> #define PLL pair<long long,long long> #define pb push_back #define fi first #define se second typedef vector<int> VI; int read(){int x = 0,f = 1;char c = getchar();while (c<‘0‘ || c>‘9‘){if (c == ‘-‘) f = -1;c = getchar();} while (c >= ‘0‘&&c <= ‘9‘){x = x * 10 + c - ‘0‘;c = getchar();}return x*f;} const double eps = 1e-9; const int maxn = 110; const int INF = 0x3f3f3f3f; const int mod = 1e9 + 7; int N; double M,K; double a[maxn]; const double PI = acos(-1.0); int main(){ while(~Sca(N)){ scanf("%lf",&M); for(int i = 1; i <= N; i ++){ scanf("%lf",&a[i]); } double ans = 0; for(int i = 1; i <= N ; i ++){ ans += M * M / 2 * sin(a[i] * PI/ 180); } printf("%.3lf\n",ans); } return 0; }
J. 0:10:46 solved by gbs
每次八位判断一下是不是97
#include <iostream> #include<stack> #include<math.h> #include<stdlib.h> #include<string.h> #include<string> #include<ctime> #include<complex> #include<stdio.h> #include<algorithm> #include<map> #include<queue> #include<deque> using namespace std; typedef long long LL; int main() { int n; unsigned int h1; while(cin >>n) { int ans =0; for (int i=0; i<n; i++) { scanf("%u",&h1); while(h1) { //cout<<(h1&255)<<endl; if ( (h1&255) ==97 ) ans++; h1>>=8; } } printf("%d\n",ans); } } /* 111 5111111 11111116 12133215 251111111 111 5 15 5 25 5 14 5 25 5 13 5 25 5 12 5 25 3 14 14 5 5 14 14 5 5 */
K. 3:15:49(-1) solved by zcz
#include<iostream> #include<cstring> #include<cstdio> using namespace std; const int maxn=5e6+5; long long mod=100000073; long long a[maxn]; long long b[maxn]; long long sum[maxn]; int main() { int c1=1,c2=1; while(c2<maxn) { a[c2]=c1; c1++; c2+=c1; } a[maxn-1]=c1; for(int i=maxn-2;i>=1;i--) { if(!a[i]) a[i]=a[i+1]; } b[1]=1; b[2]=2; b[3]=1; sum[1]=1; sum[2]=3; sum[3]=4; c1=0; for(int i=4;i<maxn;i++) { if(c1==0) c1=a[i]; //cout<<c1<<endl; // for(int j=0,k=a[i]-1;j<c1;j++,k--) // { // b[i]=(b[i]+b[i-k-1])%mod; // } b[i]=(sum[i-a[i]+c1-1]-sum[i-a[i]-1]+mod)%mod; sum[i]=(sum[i-1]+b[i])%mod; c1--; } long long a1,a2; while(scanf("%lld %lld",&a1,&a2)!=EOF) { printf("%lld %lld\n",a[a2-a1+1],b[a2-a1+1]); } return 0; }
标签：三角形 大数 lap 输出 tip pll bitset hide ini
原文地址：https://www.cnblogs.com/Hugh-Locke/p/11328979.html
