2017年山东省第八届ACM大学生程序设计竞赛 C fireworks(sdut 3895) 逆元求组合数

    距离省赛已经过去了两个多月了，然而还有一道题卡在心里迟迟没有解决，之前的博客也已经扯过了，C fireworks，还是先把题目贴出来。

   

fireworks

Time Limit: 1000MS Memory Limit: 65536KB

Problem Description

Hmz likes to play fireworks, especially when they are put regularly.
Now he puts some fireworks in a line. This time he put a trigger on each firework. With that trigger, each firework will explode and split into two parts per second, which means if a firework is currently in positionx, then in next second one part will be in position x−1 and one inx+1. They can continue spliting without limits, as Hmz likes.
Now there are n fireworks on the number axis. Hmz wants to know after T seconds, how many fireworks are there in position w?

Input

Input contains multiple test cases.
For each test case:

• The first line contains 3 integers n,T,w(n,T,|w|≤10^5)
• In next n lines, each line contains two integers xi and ci, indicating there are ci fireworks in position xi at the beginning(ci,|xi|≤10^5).

Output

For each test case, you should output the answer MOD 1000000007.

Example Input

1 2 02 22 2 20 31 2

Example Output

23

    题意是说已知有n个烟花，这些烟花会每秒变为两个，在坐标上左右再出现两个，问经过T秒后，在w位置上有多少烟花。下面n行分别表示在x位置上有c个烟花。

    在纸上简单画一下，不难发现这就是一个杨辉三角，对于求杨辉三角，这道题数据很大打表是肯定不行的，所以就要直接求组合数。当时比赛时我只了解过Lucas定理，而且之前一直都用的Lucas定理求，所以没太在意过，结果在比赛的时候一直到最后还是只得到了一堆TLE。

    现在再去看，才发现，除了Lucas之外，还有另一个方法直接求出组合数，求逆元。这个方法适用于数较小但mod较大的情况，这题mod为1000000007，所以这个题目的本意是让用求逆元的方法来做，无奈当时的我并不知道有这个方法。

    现在终于把这道心头上挂着的题做了，就像曾经说的，果然“差了一个Lucas的距离”，恩，就这样吧。

    具体的求组合数方法还是另开一贴单独整理出来吧，下面AC代码：

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const long long mod=1000000007;const long long fcnt=100005;long long fac[100005];int getfac(){    int i;    fac[0]=1;    for(i=1;i<fcnt;i++)    {        fac[i]=fac[i-1]*i%mod;    }    return 0;}long long Pow(long long a,long long b){    long long ans=1;    while(b)    {        if(b&1)        {            b--;            ans=(ans*a)%mod;        }        b>>=1;        a=(a*a)%mod;    }    return ans;}long long inv(long long a){    return Pow(a,mod-2);}long long C(long long n,long long m){    if(n<m)        return 0;    return fac[n]*inv(fac[m])%mod*inv(fac[n-m])%mod;}int main(){    long long n,T,w;    int i;    long long x,c;    long long ans;    getfac();    while(scanf("%lld%lld%lld",&n,&T,&w)!=EOF)    {        ans=0;        for(i=1;i<=n;i++)        {            scanf("%lld%lld",&x,&c);            if(abs(x-w)%2==T%2&&abs(x-w)<=T)            {                ans=(ans+(c*C(T,abs(T/2-(abs(x-w))/2)))%mod)%mod;            }        }        cout<<ans<<endl;    }    return 0;}