1.头文件
短的：
1 #include<bits/stdc++.h>
2 #define cl(a,b) memset(a,b,sizeof(a))
3 #define debug(a) cerr<<#a<<"=="<<a<<endl
4 using namespace std;
5 typedef long long ll;
6 typedef pair<int,int> pii;
7
8 const int maxn=1e5+10;
9
10 int main()
11 {
12
13 return 0;
14 }/*
15
16
17
18 */
长的：
1 #include<cstdio>
2 #include<iostream>
3 #include<algorithm>
4 #include<cmath>
5 #include<cstring>
6 #include<string>
7 #include<ctime>
8 #include<map>
9 #include<set>
10 #include<vector>
11 #include<queue>
12 #include<cstdlib>
13 #include<cassert>
14 #include<sstream>
15 #include<stack>
16 #include<list>
17 #include<bitset>
18 #define cl(a,b) memset(a,b,sizeof(a))
19 #define debug(x) cerr<<#x<<"=="<<(x)<<endl
20 using namespace std;
21 typedef long long ll;
22 typedef long double ldb;
23 typedef pair<int,int> pii;
24
25 const int inf=0x3f3f3f3f;
26 const int maxn=1e9+10;
27 const int mod=1e7+7;
28 const double eps=1e-8;
29 const double pi=acos(-1);
30
31 int dx[8]= {0,0,1,-1,1,-1,1,-1};
32 int dy[8]= {1,-1,0,0,-1,1,1,-1};
33
34 ll gcd(ll a,ll b){return a?gcd(b%a,a):b;}
35 ll powmod(ll a,ll x,ll mod){ll t=1;while(x){if(x&1)t=t*a%mod;a=a*a%mod;x>>=1;}return t;}
36 //---------------------------------------ヽ(^。^)丿
37 int main()
38 {
39
40 return 0;
41 }
42 /*
43
44
45
46 */
超神读入挂
1 namespace fastIO {
2 #define BUF_SIZE 100000
3 //fread -> read
4 bool IOerror = 0;
5 inline char nc() {
6 static char buf[BUF_SIZE], *p1 = buf + BUF_SIZE, *pend = buf + BUF_SIZE;
7 if(p1 == pend) {
8 p1 = buf;
9 pend = buf + fread(buf, 1, BUF_SIZE, stdin);
10 if(pend == p1) {
11 IOerror = 1;
12 return -1;
13 }
14 }
15 return *p1++;
16 }
17 inline bool blank(char ch) {
18 return ch == ‘ ‘ || ch == ‘\n‘ || ch == ‘\r‘ || ch == ‘\t‘;
19 }
20 inline void read(int &x) {
21 char ch;
22 while(blank(ch = nc()));
23 if(IOerror)
24 return;
25 for(x = ch - ‘0‘; (ch = nc()) >= ‘0‘ && ch <= ‘9‘; x = x * 10 + ch - ‘0‘);
26 }
27 #undef BUF_SIZE
28 };
29 using namespace fastIO;
30
31 int main()
32 {
33 while(read(n),!fastIO::IOerror)
34 {
35
36 }
37 return 0;
38 }
矩阵快速幂
1 struct matrix
2 {
3 ll a[maxn][maxn];
4 int row,col;
5 matrix():row(maxn),col(maxn){cl(a,0);}
6 matrix(int x,int y):row(x),col(y){cl(a,0);}
7 void show()
8 {
9 for(int i=0; i<row; i++)
10 {
11 for(int j=0; j<col; j++)
12 {
13 printf("%d%c",a[i][j],j==col-1?‘\n‘:‘ ‘);
14 }
15 }
16 }
17 inline ll* operator [] (int x)
18 {
19 return a[x];
20 }
21 inline matrix operator * (matrix x)
22 {
23 matrix tmp(row,col);
24 for (int i=0; i<row; i++)
25 for (int j=0; j<col; j++)
26 for (int k=0; k<row; k++)
27 tmp[i][j]=(tmp[i][j]+a[i][k]*x[k][j])%mod;
28 return tmp;
29 }
30 inline void operator *= (matrix x)
31 {
32 *this = *this * x;
33 }
34 matrix operator ^ (ll x)
35 {
36 matrix result(row,col),now=*this;
37 for(int i=0; i<row; i++)
38 {
39 result[i][i]=1;
40 }
41 while(x)
42 {
43 if(x%2) result*=now;
44 now*=now;
45 x/=2;
46 }
47 return result;
48 }
49 };
降维线段树
1 #define ls getid(l,l+r>>1)
2 #define rs getid((l+r>>1)+1,r)
3 #define lson l,m,ls
4 #define rson m+1,r,rs
5
6 typedef long long ll;
7
8 const int maxn = 1e5+10;
9
10 inline int getid(int x,int y)
11 {
12 return x+y|y!=x;
13 }
14
15 ll rm[maxn<<1],col[maxn<<1];
16
17 void pushup(int l,int r,int rt)
18 {
19 rm[rt]=rm[ls]+rm[rs];
20 }
21
22 void pushdown(int l,int r,int rt,int k)
23 {
24 if (col[rt])
25 {
26 col[ls] += col[rt];
27 col[rs] += col[rt];
28 rm[ls] += col[rt]*(k-(k>>1));
29 rm[rs] += col[rt]*(k>>1);
30 col[rt] = 0;
31 }
32 }
33
34 void build(int l,int r,int rt)
35 {
36 if(l == r)
37 {
38 scanf("%I64d",&rm[rt]);
39 return ;
40 }
41 int m=l+r>>1;
42 build(lson);
43 build(rson);
44 pushup(l,r,rt);
45 }
46
47 void update(int L,int R,int c,int l,int r,int rt)
48 {//区间更新
49 if(L<=l && r<=R)
50 {
51 col[rt]+=c;
52 rm[rt]+=c*(r-l+1);
53 return ;
54 }
55 pushdown(l,r,rt,r-l+1);
56 int m=l+r>>1;
57 if( L <= m ) update(L,R,c,lson);
58 if (R > m) update(L,R,c,rson);
59 pushup(l,r,rt);
60 }
61
62 void update(int p,int c,int l,int r,int rt)
63 {//单点更新
64 if(l==r)
65 {
66 rm[rt]+=c;
67 return ;
68 }
69 int m=l+r>>1;
70 if( p <= m ) update(p,c,lson);
71 else update(p,c,rson);
72 pushup(l,r,rt);
73 }
74
75 ll query(int L,int R,int l,int r,int rt)
76 {
77 if( L<=l && r<=R )
78 {
79 return rm[rt];
80 }
81 pushdown(l,r,rt, r-l+1);
82 int m=l+r>>1;
83 ll ret = 0;
84 if( L<=m ) ret += query(L,R,lson);
85 if( m<R ) ret += query(L,R,rson);
86 return ret;
87 }
88
89 int main()
90 {
91 int n,m;
92 scanf("%d%d",&n,&m);
93 char str[2];
94 build(1,n,1);
95 while(m--)
96 {
97 int l,r,x;
98 scanf("%s",str);
99 if(str[0]==‘Q‘)
100 {
101 scanf("%d%d",&l,&r);
102 printf("%I64d\n",query(l,r,1,n,1));
103 }
104 else
105 {
106 scanf("%d %d %d", &l, &r, &x);
107 update(l, r, x, 1, n, 1);
108 }
109 }
110 return 0;
111 }
字典树求异或值
1 typedef long long ll;
2
3 const int maxn=1e5+10;
4
5 struct t
6 {
7 ll tree[32*maxn][3];//字典树的树部分
8 ll val[32*maxn];//字典树的终止节点
9 ll cnt;//字典树的大小
10 void init() //初始化字典树
11 {
12 cl(tree,0);
13 cl(val,0);
14 cnt=1;
15 }
16 void add(ll a)
17 {
18 int x=0,now;
19 for(int i=32; i>=0; i--)
20 {
21 now=((a>>i)&1); //当前位数
22 if(tree[x][now]==0) //没有的这个节点的话 加入这个节点
23 {
24 cl(tree[cnt],0);
25 val[cnt]=0;
26 tree[x][now]=cnt++;
27 }
28 x=tree[x][now];
29 }
30 val[x]=a; //结尾标记
31 }
32 ll Search(ll a) //查找字典树中和a异或起来最大的值
33 {
34 int x=0,now;
35 for(int i=32; i>=0; i--)
36 {
37 now=!((a>>i)&1);
38 if(tree[x][now]) //如果和当前位相反的这个节点存在
39 {
40 x=tree[x][now]; //那么继续找这个节点下的子树
41 }
42 else
43 {
44 x=tree[x][!now]; //不存在就找自己这个节点下的子树
45 }
46 }
47 return val[x]; //结尾标记就是这个数字
48 }
49 } Trie;
lucas定理
1 ll fac[maxn];
2
3 ll quick_pow(ll a,ll n)
4 {
5 ll ans=1;
6 while(n)
7 {
8 if(n%2) ans=1ll*ans*a%mod;
9 a=1ll*a*a%mod;
10 n>>=1;
11 }
12 return ans;
13 }
14
15 ll inv(ll a)
16 {
17 return quick_pow(a,mod-2);
18 }
19
20 void init()
21 {
22 fac[0]=1;
23 for(int i=1;i<maxn;i++)
24 {
25 fac[i]=1ll*fac[i-1]*i%mod;
26 }
27 }
28
29 ll C(ll a,ll b)
30 {
31 ll ans=1;
32 if(a<b || b<0) return 0;
33 while(a&&b)
34 {
35 ll aa=a%mod,bb=b%mod;
36 if(aa<bb) return 0;;
37 ans = 1ll*ans*fac[aa]%mod * inv(1ll*fac[bb]*fac[aa-bb]%mod)%mod;
38 a/=mod,b/=mod;
39 }
40 return ans;
41 }
O(n)预处理Cm n
1 const int maxn=1e6+10;
2 const int mod=1e9+7;
3 ll fac[maxn],f[maxn],inv[maxn];
4
5 void init()
6 {
7 fac[0]=fac[1]=f[0]=f[1]=inv[0]=inv[1]=1;
8 for(int i=2;i<maxn;i++)
9 {
10 fac[i]=fac[i-1]*i%mod; // n！
11 ll t=mod/i,k=mod%i;
12 f[i]=(mod-t)*f[k]%mod; // n的逆元
13 inv[i]=inv[i-1]*f[i]%mod; // n！的逆元
14 }
15 }
16
17 ll C(ll n,ll m)
18 {
19 if(n<m) return 0;
20 return fac[n]*inv[m]%mod*inv[n-m]%mod;
21 }
自适应simpson积分
1 double simpson(double a,double b)
2 {
3 double c=a+(b-a)/2;
4 return (f(a)+4*f(c)+f(b))*(b-a)/6;
5 }
6
7 double asr(double a,double b,double eps,double A)
8 {
9 double c=a+(b-a)/2;
10 double L=simpson(a,c);
11 double R=simpson(c,b);
12 if(fabs(L+R-A)<=15*eps) return L+R+(L+R-A)/15.0;
13 return asr(a,c,eps/2,L)+asr(c,b,eps/2,R);
14 }
15
16 double asr(double a,double b,double eps)
17 {
18 return asr(a,b,eps,simpson(a,b));
19 }
日期公式
1 int zeller(int y,int m,int d)
2 {//计算星期几
3 if(m<=2) y--,m+=12;
4 int c=y/100;y%=100;
5 int w=((c>>2)-(c<<1)+y+(y>>2)+(13*(m+1)/5)+d-1)%7;
6 if(w<0) w+=7; return (w);
7 }
8
9 int getid(int y,int m,int d)
10 {//计算每个日期的id 可以计算两个日期之间的天数
11 if(m<3) {y--;m+=12;}
12 return 365*y+y/4-y/100+y/400+(153*m+2)/5+d;
13 }
时间： 2024-10-08 10:14:26