传送门
文章目录
- 题意:
- 思路:
题意:
给你d,modd,modd,mod,让你求能构造出如下序列aaa的个数模modmodmod:
思路:
首先可以发现aaa的长度不能超过log2dlog_2dlog2d,因为如果存在两个aaa的二进制位数相同的话,那么最高位异或之后一定会变成0,那么bbb一定小于之前的某个bbb,所以我们就看每个二进制位选或者不选来组合出来答案即可,比如当前二进制位是iii,那么他的方案就是(1<<(i+1))−(1<<i)+1(1<<(i+1))-(1<<i)+1(1<<(i+1))−(1<<i)+1,要注意到最后一位二进制的时候,需要特判一下n−(1<<i)+2n-(1<<i)+2n−(1<<i)+2,累乘起来就好辣。
// Problem: D. Dreamoon Likes Sequences
// Contest: Codeforces - Codeforces Round #631 (Div. 2) - Thanks, Denis aramis Shitov!
// URL: https://codeforces.com/contest/1330/problem/D
// Memory Limit: 256 MB
// Time Limit: 1000 ms
//
// Powered by CP Editor (https://cpeditor.org)//#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")
//#pragma GCC optimize(2)
#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
#include<map>
#include<cmath>
#include<cctype>
#include<vector>
#include<set>
#include<queue>
#include<algorithm>
#include<sstream>
#include<ctime>
#include<cstdlib>
#define X first
#define Y second
#define L (u<<1)
#define R (u<<1|1)
#define pb push_back
#define mk make_pair
#define Mid (tr[u].l+tr[u].r>>1)
#define Len(u) (tr[u].r-tr[u].l+1)
#define random(a,b) ((a)+rand()%((b)-(a)+1))
#define db puts("---")
using namespace std;//void rd_cre() { freopen("d://dp//data.txt","w",stdout); srand(time(NULL)); }
//void rd_ac() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//AC.txt","w",stdout); }
//void rd_wa() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//WA.txt","w",stdout); }typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;const int N=1000010,INF=0x3f3f3f3f;
const double eps=1e-6;int n,mod;
LL ans=1;int main()
{
// ios::sync_with_stdio(false);
// cin.tie(0);int _; cin>>_;while(_--) {cin>>n>>mod;int d=(int)log2(n);ans=1%mod;for(int i=0;i<=d;i++) {if(i<d) ans*=((1ll<<(i+1))-(1ll<<i)+1)%mod,ans%=mod;else ans*=(n-(1ll<<i)+2)%mod,ans%=mod;}ans-=1; ans+=mod; ans%=mod;cout<<ans<<endl;}return 0;
}
/**/