In the PRobability theory the following paradox called Benford's law is known: "In many lists of random numbers taken from real sources, numbers starting with digit 1 occur much more often than numbers starting with any other digit" (that's the simplest form of the law).

Having read about it on Codeforces, the Hedgehog got intrigued by the statement and wishes to thoroughly explore it. He finds the following similar problem interesting in particular: there are N random variables, the i-th of which can take any integer value from some segment [Li;Ri] (all numbers from this segment are equiprobable). It means that the value of the i-th quantity can be equal to any integer number from a given interval [Li;Ri] with probability 1?/?(Ri?-?Li?+?1).

The Hedgehog wants to know the probability of the event that the first digits of at least K% of those values will be equal to one. In other Words, let us consider some set of fixed values of these random variables and leave only the first digit (the MSD — most significant digit) of each value. Then let's count how many times the digit 1 is encountered and if it is encountered in at least K per cent of those N values, than such set of values will be called a good one. You have to find the probability that a set of values of the given random variables will be a good one.

The first line contains number N which is the number of random variables (1?≤?N?≤?1000). Then follow N lines containing pairs of numbers Li,?Ri, each of whom is a description of a random variable. It is guaranteed that 1?≤?Li?≤?Ri?≤?1018.

The last line contains an integer K (0?≤?K?≤?100).

All the numbers in the input file are integers.

Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cin (also you may use %I64d).

Print the required probability. Print the fractional number with such a precision that the relative or absolute error of the result won't exceed 10?-?9.

11 250output
0.500000000000000input
21 29 1150output
0.833333333333333
題意：
給出n個數，然后給出n行，每行li，ri，代表第i個數在區間[li,ri]中，求一個概率使得這n個數中有K%的數是1開頭的。
題解：
首先，可以先求出每個區間以1開頭的數有多少個，即每個區間滿足條件的概率pi，要使得有K%的區間滿足條件，這個該怎么求呢？
我一開始以為是概率方面的題，想了一堆容斥什么的，發現都不可做，最后看了官方題解。
題解是用概率dp做的（妙啊！），d[i][j]表示前i個數有j個滿足條件的概率。
#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>#include<vector>#include<queue>#include<stack>using namespace std;#define rep(i,a,n) for (int i=a;i<n;i++)#define per(i,a,n) for (int i=n-1;i>=a;i--)#define pb push_back#define fi first#define se secondtypedef vector<int> VI;typedef long long ll;typedef pair<int,int> PII;const int inf=0x3fffffff;const ll mod=1000000007;const int maxn=1000+10;ll f[30];ll pow(int x){    ll t=10,ans=1;    while(x)    {        if(x&1) ans*=t;        t=t*t;        x/=2;    }    return ans;}ll cal(ll x){    int cnt=0;    ll t=x;    ll las=0;    while(t)    {        las=t;        cnt++;        t/=10;    }    if(las!=1) return f[cnt];    else return x-pow(cnt-1)+1+f[cnt-1];}double p[maxn];double d[maxn][maxn];int main(){    f[1]=1;    ll q=1;    rep(i,2,20)    {        q*=10;        f[i]=f[i-1]+q;    }    int n;    scanf("%d",&n);    rep(i,1,n+1)    {        ll l,r;        scanf("%lld%lld",&l,&r);        ll ans=cal(r)-cal(l-1);        p[i]=1.0*ans/(r-l+1);    }    int k;    scanf("%d",&k);    d[0][0]=1;    rep(i,1,n+1) rep(j,0,i+1)    {        if(j>0)        d[i][j]=d[i-1][j-1]*p[i]+d[i-1][j]*(1-p[i]);        else d[i][j]=d[i-1][j]*(1-p[i]);    }    double kk=1.0*k/100;    double ans=0;    rep(i,0,n+1)    {        double t=1.0*i/n;        if(t>=kk) ans+=d[n][i];    }    printf("%.9lf/n",ans);    return 0;}