标签：链接 std void cti efi line extern circle set
规律看两两之间的差
4^n+3出现的规律就是把一个数的二进制表示的情况下通过删除或插入或替代的方法从n到n+1的次数
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BCEF
C:
#include<cstdio> #include<iostream> #include<cstring> using namespace std; int main() { int t,x; scanf("%d",&t); while(t--){ cin>>x; printf("1"); for(int i=2;i<=233;i++) { printf("0000001"); } printf("\n"); } return 0; }
B：
#include <iostream> #include <cstdio> using namespace std; int mod=1e9+7; typedef long long ll; long long exgcd(long long a,long long b,long long &x,long long &y) { if(a==0&&b==0) return -1;//无最大公约数 if(b==0){x=1;y=0;return a;} long long d=exgcd(b,a%b,y,x); y-=a/b*x; return d; } long long inv(long long a,long long n) { long long x,y; long long d=exgcd(a,n,x,y); if(d==1) return (x%n+n)%n; else return -1; } ll pow(int x,int n) //注意n等于0的时候 { ll a=x; ll ans=1; while (n) { if (n&1) ans=ans*a%mod; a=a*a%mod; n>>=1;//效果跟n/=2；一样 } return ans; } int main() { int T; scanf("%d",&T); while(T--) { ll p,q,k; ll x,y; scanf("%lld%lld%lld",&p,&q,&k); if(p>q) { swap(q,p); } x=pow(q-2*p,k); y=pow(q,k); x=(x+y)%mod; y=(2*y)%mod; printf("%lld\n",(x*inv(y,mod))%mod); } }
#include<cstdio> #include<cstring> #include<algorithm> #include<iostream> #include<cmath> using namespace std; const int mod = 1e9+7; long long _pow(long long a,long long b) { long long res = 1; while(b) { if(b&1) res = res*a%mod; a=a*a%mod; b>>=1; } return res; } long long exgcd(long long a,long long b,long long &x,long long &y) { if(a==0&&b==0) return -1;//无最大公约数 if(b==0){x=1;y=0;return a;} long long d=exgcd(b,a%b,y,x); y-=a/b*x; return d; } long long inv(long long a,long long n) { long long x,y; long long d=exgcd(a,n,x,y); if(d==1) return (x%n+n)%n; else return -1; } int main() { int t; scanf("%d",&t); while(t--) { int p,q,k; scanf("%d%d%d",&p,&q,&k); if(p>q){ int temp=p; p=q; q=temp; } long long x=_pow(q-2*p,k); long long y=_pow(q,k); x=y+x; y=2*y; printf("%lld\n",x*inv(y,mod)%mod); } return 0; }
E:
#include <bits/stdc++.h> using namespace std; typedef long long ll; const ll mod=1000000007; ll sum[100]; ll la[100]; ll two[100]; int main() { sum[0]=1; la[0]=1; two[0]=1; for(int i=1; i<=64; i++) { two[i]=two[i-1]*2ll; sum[i]=(sum[i-1]*2%mod+la[i-1]*3%mod)%mod; la[i]=la[i-1]*4%mod; } ll n; while(~scanf("%lld",&n)) { n--; ll pos; for(ll i=62; i>=0; i--) { if(n>=two[i]) { pos=i; break; } } ll ans=n%mod; // cout<<pos<<endl; pos=(1ll<<pos); n-=pos; ans+=((1ll*pos/2ll)%mod)*((pos-1ll)%mod); ans%=mod; for(ll i=0; i<63; i++) { if(n&(1ll<<i)) ans+=sum[i]; ans%=mod; } printf("%lld\n",ans%mod); } }
#include<iostream> #include<cstdio> #include<cstring> #include<string> using namespace std; const int MAXN = 30010; string lib[MAXN]; int tot; int cmpa(string &str, string &a){ for(int i = 0; i < a.length(); ++i){ if(str[i] != a[i]){ return 0; } } return 1; } int cmpd(string &str, string &d){ for(int i = 1; i <= d.length(); ++i){ if(str[str.length() - i] != d[d.length() - i]){ return 0; } } return 1; } int cmpbc(string &str, string &b, string &c){ int mid = str.length() / 2; for(int i = 0; i < c.length(); ++i){ if(str[mid + i] != c[i]){ return 0; } } for(int i = 1; i <= b.length(); ++i){ if(str[mid - i] != b[b.length() - i]){ return 0; } } return 1; } int main(){ ios::sync_with_stdio(false); int T, q; for(cin >> T; T && cin >> q; --T){ tot = 0; int ctrl; for (int k = 0; k < q; ++k) { cin >> ctrl; if (ctrl == 1) { cin >> lib[tot++]; } else { string a, b, c, d; cin >> a >> b >> c >> d; int maxLen = a.length() + b.length(); if (c.length() + d.length() > maxLen) maxLen = c.length() + d.length(); maxLen *= 2; int cnt = 0; for (int i = 0; i < tot; ++i) { if (lib[i].length() >= maxLen && lib[i].length() % 2 == 0) { if (cmpa(lib[i], a) && cmpd(lib[i], d) && cmpbc(lib[i], b, c)) { ++cnt; } } } cout << cnt << endl; } } } return 0; }
F:【分析】：
http://www.docin.com/p-385138324.html?qq-pf-to=pcqq.group加个费马小定理 乘逆元就可以了
#include<iostream> #include<cstdio> #include<string> #include<cstring> #include<cstdlib> #include<sstream> #include<fstream> #include<vector> #include<list> #include<deque> #include<stack> #include<queue> #include<map> #include<set> #include<cmath> #include<utility> #include<numeric> #include<iterator> #include<algorithm> #include<functional> #include<ctime> #include<cassert> using std::cin; using std::cout; using std::endl; typedef long long ll; typedef unsigned long long ull; typedef std::pair<int,int> P; #define FOR(i,init,len) for(int i=(init);i<(len);++i) #define For(i,init,len) for(int i=(init);i<=(len);++i) #define fi first #define se second #define pb push_back #define is insert namespace IO { inline char getchar() { static const int BUFSIZE=5201314; static char buf[BUFSIZE],*begin,*end; if(begin==end) { begin=buf; end=buf+fread(buf,1,BUFSIZE,stdin); if(begin==end) return -1; } return *begin++; } } inline void read(int &in) { int c,symbol=1; while(isspace(c=IO::getchar())); if(c==‘-‘) { in=0;symbol=-1; } else in=c-‘0‘; while(isdigit(c=IO::getchar())) { in*=10;in+=c-‘0‘; } in*=symbol; } inline int read() { static int x;read(x);return x; } ll gcd(ll a,ll b) { return b?gcd(b,a%b):a; } ll lcm(ll a,ll b) { return a/gcd(a,b)*b; } const ll mod=998244353LL; const int maxm=1e4+10; ll qpow(ll a,ll b) { ll ans=1; while(b) { if(b&1) ans=ans*a%mod; a=a*a%mod; b>>=1; } return ans; } ll f[maxm],invf[maxm]; ll n,m; int main() { #ifdef MengLan int Beginning=clock(); //freopen("in","r",stdin); //freopen("out","w",stdout); #endif // MengLan f[0]=1; FOR(i,1,maxm) f[i]=f[i-1]*i%mod; invf[maxm-1]=qpow(f[maxm-1],mod-2); for(int i=maxm-2;i>=0;--i) invf[i]=invf[i+1]*(i+1)%mod; while(cin>>n>>m) { if((n&1)^(m&1)) { puts("0");continue; } if(m>n) { puts("0");continue; } if(m==0) { puts("998244352");continue; } ll ans=1; for(ll i=n-m+2;i<=n+m-2;i+=2) { ans*=i;ans%=mod; } ans=ans*n%mod; ll k=1; For(i,1,m) { k=k*i%mod; } ll x=qpow(k,mod-2); ans=ans*x%mod; if(((n-m)/2)&1) ans=(mod-ans)%mod; cout<<ans<<endl; } #ifdef MengLan printf("Time: %d\n",clock()-Beginning); #endif // MengLan return 0; }
标签：链接 std void cti efi line extern circle set
原文地址：http://www.cnblogs.com/Roni-i/p/7531928.html
