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Security Research, Algorithm and Data Structure

51Nod 1799 二分答案

Problem

lyk最近在研究二分答案类的问题。

对于一个有n个互不相同的数且从小到大的正整数数列a(其中最大值不超过n),若要找一个在a中出现过的数字m,一个正确的二分程序是这样子的:

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l=1; r=n; mid=(l+r)/2;
while (l<=r)
{
if (a[mid]<=m)
l=mid+1;
else
r=mid-1;

mid=(l+r)/2;
}

最终a[r]一定等于m。

但是这个和谐的程序被熊孩子打乱了。

熊孩子在一开始就将a数组打乱顺序。(共有n!种可能)

lyk想知道最终r=k的期望。

由于小数点非常麻烦,所以你只需输出将答案乘以n!后对1000000007取模就可以了。

在样例中,共有2个数,被熊孩子打乱后的数列共有两种可能(1,2)或者(2,1),其中(1,2)经过上述操作后r=1,(2,1)经过上述操作后r=0。r=k的期望为0.5,0.5*2!=1,所以输出1。

Solution

模拟二分找到k,记录关键节点,计算排列,非关键节点阶乘。

Code

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#include<stdio.h>
#include<set>
#include<iostream>
#include<stack>
#include<cstring>
#include<cmath>
#include<vector>
#include<map>
#include<queue>
#include<algorithm>
typedef long long ll;
typedef long double ld;
typedef double db;
#define io_opt ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
using namespace std;
const int mod=1e9+7;
inline int mo(ll a,int p){
return a>=p?a%p:a;
}
inline int rd() {
int x = 0, f = 1;
char ch;
while (ch < '0' || ch > '9') {
if (ch == '-')f = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9') {
x = x * 10 + ch - '0';
ch = getchar();
}
return f * x;
}
inline ll gcd(ll x, ll y){
return y==0?x:gcd(y,x%y);
}
inline ll speed(ll a,ll b){
ll cur=a,anss=1;
while(b){
if(b&1) anss=anss*cur;
cur=cur*cur;
b>>=1;
}
return anss;
}
const int MAXN=1e5;
bool ipr[MAXN+20];
int cnt,pri[MAXN/5];
void prime(){//埃式筛法
int N=sqrt(MAXN)+0.5,mul;
memset(ipr,true,sizeof(ipr));
ipr[1]=false;
for(int i=2;i<=N;i++){
if(ipr[i]==true){
i==2?mul=1:mul=2;
for(int j=i*i;j<=MAXN;j+=i*mul){
ipr[j]=false;
}
}
}
for(int i=2;i<=MAXN;i++){
if(ipr[i]==true){
pri[++cnt]=i;
}
}
}
int f[120]={1,682498929,491101308,76479948,723816384,67347853,27368307,625544428,199888908,888050723,
927880474,281863274,661224977,623534362,970055531,261384175,195888993,66404266,547665832,109838563,
933245637,724691727,368925948,268838846,136026497,112390913,135498044,217544623,419363534,500780548,
668123525,128487469,30977140,522049725,309058615,386027524,189239124,148528617,940567523,917084264,
429277690,996164327,358655417,568392357,780072518,462639908,275105629,909210595,99199382,703397904,
733333339,97830135,608823837,256141983,141827977,696628828,637939935,811575797,848924691,131772368,
724464507,272814771,326159309,456152084,903466878,92255682,769795511,373745190,606241871,825871994,
957939114,435887178,852304035,663307737,375297772,217598709,624148346,671734977,624500515,748510389,
203191898,423951674,629786193,672850561,814362881,823845496,116667533,256473217,627655552,245795606,
586445753,172114298,193781724,778983779,83868974,315103615,965785236,492741665,377329025,847549272,
698611116};
const int con=1e7;
int n,m,k;
ll ans=1;
ll fun(ll x){
ll ret=f[x/con];
ll t=x/con;
for(int i=t*con+1;i<=x;i++){
ret=mo(ret*i,mod);
}
return ret;
}
int main(){
//io_opt;
scanf("%d%d%d",&n,&m,&k);
int a=0,b=0;
int l=1,r=n,mid=(l+r)/2;
while (l<=r)
{
if (mid<=k)
l=mid+1,a++;
else
r=mid-1,b++;

mid=(l+r)/2;
}
for(int i=m;i>=m-a+1;i--){
ans=mo(ans*i,mod);
}
for(int i=n-m;i>=n-m-b+1;i--){
ans=mo(ans*i,mod);
}
ans=ans*fun(n-a-b)%mod;
printf("%lld\n",ans);
return 0;
}

  • 本文作者: CCWUCMCTS
  • 本文链接: https://ccwucmcts.github.io/posts/17706/
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