[HDU-3487]Play with Chain
F – Play with Chain Time Limit:2000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u Submit Status Description YaoYao is fon…
Read More »[ POJ2187]Beauty Contest——旋转卡壳初步
题目传送门—-> Beauty Contest Description Bessie, Farmer John’s prize cow, has just won first place in a bovine beauty contest, earning the title ‘Miss Cow World’. As a result, Bessie will make a tour …
Read More »Splay[转]
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#include <iostream> #include <cstdio> #include <cstring> #include <string> #include <cmath> #include <algorithm> #include <cstdlib> #include <vector> using namespace std; #define ll ch[x][0] #define rr ch[x][1] #define KT (ch[ch[rt][1]][0]) const int maxn = 300100; int a[maxn], flage[maxn]; int n, m; struct SplayTree{ ///基本数据定义 int ch[maxn][2]; int sz[maxn], pre[maxn]; int rt, tot; ///题意数据定义 int val[maxn], flg[maxn], GCD[maxn][2]; ///Splay树的基本旋转操作函数 void Rotate(int x, int f) { int y = pre[x]; // down(y); down(x);/// ch[y][!f] = ch[x][f]; pre[ch[x][f]] = y; pre[x] = pre[y]; if (pre[x]) ch[pre[y]][ch[pre[y]][1] == y] = x; ch[x][f] = y; pre[y] = x; up(y);/// } void Splay(int x, int goal) { // down(x);/// while (pre[x] != goal) { if (pre[pre[x]] == goal) Rotate(x, ch[pre[x]][0] == x); else { int y = pre[x], z = pre[y]; int f = ( ch[z][0] == y ); if (ch[y][f] == x) Rotate(x, !f), Rotate(x, f); else Rotate(y, f), Rotate(x, f); } } up(x);/// if (goal == 0) rt = x; } void RTO(int k, int goal) { int x = rt; while (sz[ll] + 1 != k) { if (k < sz[ll] + 1) x = ll; else { k -= (sz[ll] + 1); x = rr; } } Splay(x, goal); } ///Splay树的生成函数 void newnode(int &x, int c, int state, int f)/// { x = ++tot; ll = rr = 0; sz[x] = 1; pre[x] = f; val[x] = c; flg[x] = state; GCD[x][state] = c; GCD[x][!state] = -1; } void build(int &x, int l, int r, int f)/// { if (l > r) return ; int m = (l + r) >> 1; newnode(x, a[m], flage[m], f); build(ll, l, m - 1, x); build(rr, m + 1, r, x); up(x); } void Init(int n) { ch[0][0] = ch[0][1] = pre[0] = sz[0] = 0; GCD[0][0] = GCD[0][1] = -1; flg[0] = val[0] = 0;///???? rt = tot = 0; newnode(rt, -1, 0, 0); newnode(ch[rt][1], -1, 0, rt); build(KT, 1, n, ch[rt][1]);/// up(ch[rt][1]); up(rt);/// } ///区间操作相关up(), down() void up(int x)///!!! { sz[x] = sz[ll] + sz[rr] + 1;///!!! GCD[x][flg[x]] = val[x];///!!!! GCD[x][!flg[x]] = -1; for (int i = 0; i < 2; i++) { if (GCD[ll][i] != -1) { if (GCD[x][i] != -1) GCD[x][i] = __gcd(GCD[x][i], GCD[ll][i]); else GCD[x][i] = GCD[ll][i]; } if (GCD[rr][i] != -1) { if (GCD[x][i] != -1) GCD[x][i] = __gcd(GCD[x][i], GCD[rr][i]); else GCD[x][i] = GCD[rr][i]; } } } ///区间查询 void solve_Q(int L, int R, int state) { RTO(L, 0); RTO(R + 2, rt); printf("%d\n", GCD[KT][state]); } ///插入一个节点 void solve_I(int L, int value, int state) { RTO(L + 1, 0); RTO(L + 2, rt); int x; newnode(x, value, state, ch[rt][1]);///!!! KT = x; up(ch[rt][1]); up(rt); } ///删除一个节点 void solve_D(int L) { RTO(L, 0); RTO(L + 2, rt); KT = 0; up(ch[rt][1]); up(rt); } ///更改节点信息 void solve_R(int L) { RTO(L + 1, 0); flg[rt] ^= 1; up(rt); } void solve_M(int L, int value) { RTO(L + 1, 0); val[rt] = value; up(rt); } ///题意相关函数 void solve(int m) { char op; int L, R, state, value; while (m--) { scanf(" %c", &op); if (op == 'Q') { scanf("%d%d%d", &L, &R, &state); solve_Q(L, R, state); } else if (op == 'I') { scanf("%d%d%d", &L, &value, &state); solve_I(L, value, state); } else if (op == 'D') { scanf("%d", &L); solve_D(L); } else if (op == 'R') { scanf("%d", &L); solve_R(L); } else if (op == 'M') { scanf("%d%d", &L, &value); solve_M(L, value); } } } }sp; int main() { while (cin >> n >> m) { for (int i = 1; i <= n; i++) scanf("%d%d", &a[i], &flage[i]); sp.Init(n); sp.solve(m); } } |
感觉很不错的样子，转过来看看。
Read More »插值求多项式系数
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#include <iostream> #include <cstring> #include <algorithm> #include <cstring> #include <map> #include <vector> #include <cmath> using namespace std; const int maxm=100; const int mod=1e9+7; const double EPS=1e-8; typedef vector<double> vec; typedef vector<vec> mat; vec gauss_jordan(const mat&A,const vec&b) { int n=A.size(); mat B(n,vec(n+1)); for (int i=0;i<n;i++) { for (int j=0;j<n;j++) B[i][j]=A[i][j]; } for (int i=0;i<n;i++) B[i][n]=b[i]; for (int i=0;i<n;i++) { int pivot=i; for (int j=i;j<n;j++) { if (abs(B[j][i])>abs(B[pivot][i])) pivot=j; } swap(B[i],B[pivot]); if (abs(B[i][i])<EPS) return vec(); for (int j=i+1;j<=n;j++) B[i][j]/=B[i][i]; for (int j=0;j<n;j++) { if (i!=j) { for (int k=i+1;k<=n;k++) B[j][k]-=B[j][i]*B[i][k]; } } } vec x(n); for (int i=0;i<n;i++) { x[i]=B[i][n]; } return x; } mat cheng(mat m,mat mm) { int n=m.size(); mat res(n,vec(n)); for (int i=0;i<n;i++) { for (int j=0;j<n;j++) { for (int k=0;k<n;k++) { res[i][k]=res[i][k]+m[i][j]*mm[j][k]; } } } return res; } mat qpow(mat m,long long b) { int n=m.size(); mat ret(n,vec(n)); for (int i=0;i<n;i++) { ret[i][i]=1; } while(b) { if (b&1) ret=cheng(ret,m); m=cheng(m,m); b/=2; } return ret; } long long a[maxm]; long long sum[maxm];//插值的公式值 map<int,int> f; long long get(int n) { int crt=1; while(f[crt]) { crt++; } return crt; } const int maxn=50;//项数 mat A(maxn,vec(maxn)); vec b(maxn); int main() { a[1]=1; a[2]=2; f[1]=1; sum[1]=1; sum[2]=3; for (int i=3;i<maxm;i++) { if (i%2) { a[i]=2*a[i-1]; } else { int crt=get(i); a[i]=a[i-1]+crt; } for (int j=1;j<i;j++) { f[a[i]-a[j]]=1; } a[i]%=mod; sum[i]=(a[i]+sum[i-1]) % mod; } for (int i=maxn;i<=maxn+maxn-1;i++) { b[i-maxn]=sum[i]; } for (int i=0;i<maxn;i++) { for (int j=0;j<maxn-1;j++) { A[i][j]=sum[i+j+1]; } A[i][maxn-1]=1; } vec x=gauss_jordan(A,b); for (int i=0;i<maxn;i++) cout<<x[i]<<endl; mat B(maxn,vec(maxn)); for (int i=0;i<maxn-1;i++) { B[0][i]=x[maxn-2-i]; } B[0][maxn-1]=x[maxn-1]; for (int i=1;i<maxn;i++) { for (int j=0;j<maxn;j++) { B[i][j]=0; } if (i<maxn-1) B[i][i-1]=1; else B[i][i]=1; } for (int i=0;i<maxn;i++) { for (int j=0;j<maxn;j++) { cout<<B[i][j]<<" "; } cout<<endl; } for (int i=maxn;i<100;i++) { mat ret=qpow(B,i-maxn+1); double ans=0; for (int j=0;j<maxn-1;j++) { ans+=sum[maxn-1-j]*ret[0][j]; } ans+=ret[0][maxn-1]; cout<<ans<<" "<<sum[i]<<endl; } return 0; } |
数论模板（一）
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//最大公约数 long long gcd(long long a,long long b) { return b==0?a:gcd(b,a%b); } //大整数相乘取模 long long multi(long long a,long long b,long long mod) { long long ret=0; while(b) { if (b&1) ret=(ret+a) %mod; a=(a<<1)%mod; b=b>>1; } return ret; } //快速幂1 long long quickpow(long long a,long long b,long long mod) { long long ret=1; while(b) { if (b&1) ret=multi(ret,a,mod); a=multi(a,a,mod); b>>=1; } return ret; } //快速幂2 long long quick_pow(long long a,long long b) { long long res=1; while(b>0) { if (b&1) res=(res*a)%p; a=(a*a)%p; b>>=1; } return res; } //欧拉函数 long long eular(long long n) { long long ret=1,i; for (i=2;i*i<=n;i++) { if (n%i==0) { n=n/i; ret*=i-1; while(n%i==0) { n=n/i; ret*=i; } } } if (n>1) ret*=n-1; return ret; } //预处理maxn范围内的欧拉函数 void euler_pre() { for (int i = 0; i <= maxn; ++i) euler[i] = i; for (int i = 2; i <= maxn; ++i) { if (euler[i] == i) { for (int j = i; j <= maxn; j += i) euler[j] = euler[j] / i * (i - 1); } } } //求一个数的所有约数，返回约数个数，约数存在c数组 int getdivision(int n) { int i; int length=0; for(i=1;i*i<n;i++){ if(n%i == 0){ c[length ++] = i; c[length ++] = n/i; } } if(i*i == n) c[length ++] = i; return length; } //矩阵（n×n）,p 为模数 struct Mat { long long mat[maxn][maxn]; long long n; Mat operator * (Mat b) { Mat c; c.n=n; memset(c.mat, 0, sizeof(c.mat)); int i, j, k; for(k = 0; k < n; ++k) { for(i = 0; i < n; ++i) { for(j = 0; j < n; ++j) { c.mat[i][j] =(c.mat[i][j]+(mat[i][k] * b.mat[k][j]) %p)%p; } } } return c; } }; //矩阵快速幂 Mat quick_mat(Mat a,long long k，long long p;) { Mat c; c.n=a.n; int i, j; for(i = 0; i < a.n; ++i) for(j = 0; j < a.n; ++j) c.mat[i][j] = (i == j); for(; k; k >>= 1) { if(k&1) c = c*a; a = a*a; } return c; } //高斯消元 typedef vector<double> vec; typedef vector<vec> mat; mat A; vec gauss_jordan(int n) { for (int c=0,r=0;c<n;++c,++r) { int pivot =r; for (int i=r+1;i<=n;++i) { if (abs(A[i][c])>abs(A[pivot][c])) pivot = i; } swap(A[r],A[pivot]); for (int j=n;j>=c;--j) A[r][j]/=A[r][c]; for (int i=0;i<=n;i++) { if (i!=r) { for (int k=c+1;k<=n;k++) { A[i][k]-=A[i][c]*A[r][k]; } } } } vec x(n); for (int i=0;i<n;i++) x[i]=A[i][n]; return x; } //组合数取模，Lucas定理，p为模数 LL C(LL n, LL m) { if(m > n) return 0; LL ans = 1; for(int i=1; i<=m; i++) { LL a = (n + i - m) % p; LL b = i % p; ans = ans * (a * quick_mod(b, p-2) % p) % p; } return ans; } LL Lucas(LL n, LL m) { if(m == 0) return 1; return C(n % p, m % p) * Lucas(n / p, m / p) % p; } //莫比乌斯函数 const int maxn=1000000; bool check[maxn+10]; int prime[maxn+10]; int mu[maxn+10]; void moblus() { memset(check,false,sizeof check); mu[1]=1; int tot=0; for (int i=2;i<=maxn;i++) { if (!check[i]) { prime[tot++]=i; mu[i]=-1; } for (int j=0;j<tot;j++) { if (i*prime[j]>maxn) break; check[i*prime[j]]=true; if (i%prime[j]==0) { mu[i*prime[j]]=0; break; } else { mu[i*prime[j]]=-mu[i]; } } } } //扩展欧几里德定理 typedef long long LL; void extend_Euclid(LL a, LL b, LL &x, LL &y) { if(b == 0) { x = 1; y = 0; return; } extend_Euclid(b, a % b,x, y); LL tmp = x; x = y; y = tmp - a / b * y; } //中国剩余定理 typedef long long LL; LL RemindChina(LL a[],LL m[],int k) { LL M = 1; LL ans = 0; for(int i=0; i<k; i++) M *= m[i]; for(int i=0; i<k; i++) { LL x, y; LL Mi = M / m[i]; extend_Euclid(Mi, m[i], x, y); ans = (ans + Mi * x * a[i]) % M; } if(ans < 0) ans += M; return ans; } |
Homework of History
题目： Homework of History The History teacher gives Nobita a homework, explore the history of 1000 yeas ago.So Nobita and Doraemon go back to 1000 years ago through time machine.But there are something …
Read More »Homework of Politics
题目： Homework of Politics After the Politics class, the teacher gives the student a homework.But Nobita and Takeshi don’t like this homework at all.So they decide to play a game, and the person who los…
Read More »splay?
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可能我写了假的splay<img src="http://zhuyeye.cn/wp-content/plugins/kindeditor-for-wordpress/plugins/emoticons/images/17.gif" alt="" border="0" /> |
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#include <iostream> #include <cstdio> #include <cstring> using namespace std; const int maxn=10+2; int f[maxn],ch[maxn][2],key[maxn],cnt[maxn],Size[maxn]; int root,sz; void Clear(int x) { ch[x][0]=ch[x][1]=f[x]=key[x]=cnt[x]=Size[x]=0; return; } int get(int x) { return ch[f[x]][1]==x; } void update(int x) { if (x) { Size[x]=cnt[x]; if (ch[x][0]) Size[x]+=Size[ch[x][0]]; if (ch[x][1]) Size[x]+=Size[ch[x][1]]; } } void Rotate(int x) { int old=f[x],oldf=f[old],which=get(x); ch[old][which]=ch[x][which^1]; f[ch[old][which]]=old; f[0]=0; f[old]=x; ch[x][which^1]=old; f[x]=oldf; if (oldf) { ch[oldf][ch[oldf][1]==old]=x; } update(old); update(x); } void splay(int x) { for (int fa;(fa=f[x]);Rotate(x)) { if (f[fa]) { Rotate((get(x)==get(fa))?fa:x); } } root=x; } void Insert(int v) { if (root==0) { sz++; ch[sz][0]=ch[sz][1]=f[sz]=0; cnt[sz]=1; key[sz]=v; Size[sz]=1; root=sz; return; } int now=root,fa=0; while (true) { if (key[now]==v) { cnt[now]++; update(now); update(fa); splay(now); break; } fa=now; now=ch[now][key[now]<v]; if (now==0) { sz++; key[sz]=v; cnt[sz]=Size[sz]=1; ch[sz][0]=ch[sz][1]=0; f[sz]=fa; ch[fa][key[fa]<v]=sz; update(fa); splay(sz); break; } } } int Find(int v) { int ans=0,now=root; while(true) { if (v<key[now]) { now=ch[now][0]; } else { ans+=(ch[now][0]?Size[ch[now][0]]:0); if (key[now]==v) { splay(now); return ans+1; } ans+=cnt[now]; now=ch[now][1]; } if (now==0) return -1; } } int findx(int x) { if (x>sz) return -1; int now=root; while (true) { if (ch[now][0]&&Size[ch[now][0]]>=x) { now=ch[now][0]; } else { int temp=(ch[now][0]?Size[ch[now][0]]:0)+cnt[now]; if (x<=temp) return key[now]; x-=temp;now=ch[now][1]; } } } int pre() { int now =ch[root][0]; while(ch[now][1]) now=ch[now][1]; return now; } int next() { int now=ch[root][1]; while(ch[now][0]) now=ch[now][0]; return now; } void del(int x){ int whatever=Find(x); if (cnt[root]>1) {cnt[root]–;return;} //Only One Point if (!ch[root][0]&&!ch[root][1]) {Clear(root);root=0;return;} //Only One Child if (!ch[root][0]){ int oldroot=root;root=ch[root][1];f[root]=0;Clear(oldroot);return; } else if (!ch[root][1]){ int oldroot=root;root=ch[root][0];f[root]=0;Clear(oldroot);return; } //Two Children int leftbig=pre(),oldroot=root; splay(leftbig); f[ch[oldroot][1]]=root; ch[root][1]=ch[oldroot][1]; Clear(oldroot); update(root); return; } int main() { int n; root=0; sz=0; scanf(“%d”,&n); for (int i=0;i<n;i++) { int opt,x; scanf(“%d%d”,&opt,&x); switch (opt) { case 1: { Insert(x); break; } case 2:{ del(x); break; } case 3:{ printf(“%dn”,Find(x)); break; } case 4:{ printf(“%dn”,findx(x)); break; } case 5:{ Insert(x); printf(“%dn”,key[pre()]); del(x); break; } case 6:{ Insert(x); printf(“%dn”,key[next()]); del(x); break; } } } return 0; } |