山东省第八届acm省赛C题 fireworks

题目来戳呀

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 position x, then in next second one part will be in position x−1 and one in x+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 0
2 2
2 2 2
0 3
1 2

Example Output

2
3

题意：
放烟花，在xi位置上有ci个烟花，1秒钟一个分成两个,前一个后一个，此位置上的就没了。
一共有n个位置，问t秒后，在w位置上有多少个烟花。

想法：
写几组数据，用图的形式展开来（不要紧巴巴的真的就在数轴写啊+_+鬼能看出来啊）
某个位置上的图写出来发现类似杨辉三角，但是中间是加了0的
0 0 0 0 0 1 0 0 0 0 0
0 0 0 0 1 0 1 0 0 0 0
0 0 0 1 0 2 0 1 0 0 0
0 0 1 0 3 0 3 0 1 0 0
0 1 0 4 0 6 0 4 0 1 0
当所在的位置的奇偶性和时间的奇偶性相同时，发现符合杨辉三角。
符合杨辉三角的部分，即求组合数，C(n,m)=n!/m!(n-m)!，再对1e9+7取模；
因为取模之前的结果太大了，所以我们通过乘法逆元的方法来求，即除以一个数等于乘以这个数的逆元再取模，这位的逆元写的贼好
乘法逆元这里选择用快速幂的方法来求。
不符合杨辉三角的就是0。

#include<cstdio>#include<algorithm>#include<iostream>#include<cstring>#include<cmath>using namespace std;typedef long long ll;const int mod=1e9+7;ll arr[100010];void init()//求阶乘{    arr[0]=arr[1]=1;    for(int i=2;i<=1e5;++i)        arr[i]=(arr[i-1]*i)%mod;}ll PowerMod(ll a, ll b)//快速幂{    ll ans=1;    a=a%mod;    while(b>0)    {        if(b&1)            ans=(ans*a)%mod;        b>>=1;        a=(a*a)%mod;    }    return ans;}ll cn(ll n,ll m){    return ((arr[n]*PowerMod(arr[n-m],mod-2))%mod*PowerMod(arr[m],mod-2))%mod;}int main(){    init();    int n,t,w;    while(cin>>n>>t>>w)    {        ll ans=0;        while(n--)        {            ll ci,xi;            cin>>xi>>ci;            ll k=abs(w-xi);//求绝对值            if((k&1)==(t&1)&&k<=t)//奇偶性相同就符合杨辉三角并且在时间范围内                ans=(ans+(ci*cn(t,(k+t)/2))%mod)%mod;        }         cout<<ans<<endl;    }    return 0;}

ps： 哇卡哇卡不出来了QAQ