Fireworks（山东省第8届ACM省赛）逆元，组合数

fireworks

Time Limit: 1000MS Memory Limit: 65536KB

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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

 3 

 2

这题的大概意思就是定义一种能无限炸裂的烟花，每秒钟烟花炸裂后原位置什么都不会剩下，而和他相邻的两个位置会出现两个子烟花，然后让你求t秒后位置w的烟花个数。题目又给定了n组烟花，所以要求的是n组烟花在位置w的子烟花个数。

 仔细观察会发现，烟花分裂的规律符合杨辉三角（去掉所有为0的点），求组合数，由于组合数太大，所以用逆元来代替除法取模。
本题使用快速幂跟费马小定理求逆元。

#include
#include
using namespace std;
typedef long long ll;
const int maxn = 1299709 + 10;
const int maxt = 100200;
const int inf = 0x3f3f3f3f;
const ll INF = 0x7f7f7f7f7f7f7f7f;
const int mod = 1e9 + 7;
const double pi = acos(-1.0);
const double eps = 1e-8;
int a[maxn];
ll C[maxn];
ll quick_mod(ll x, int n){
ll ret = 1;
while(n){
if(n & 1) ret = ret * x % mod;
x = x * x %mod;
n >>= 1;
}
return ret;
}
void init(int t){
C[0] = 1;
for(int i = 1; i <= t; ++i){
C[i] = C[i - 1] * (t - i + 1) % mod * quick_mod(i, mod - 2) % mod;
}
}
ll solve(ll x, int w, int l){
int d = abs(x - w);
ll sum = 0;
if(l & 1){
if(d < l && d % 2 == 0)
sum += C[l / 2 - d
a67e
/ 2];
}
else{
if(d < l && d & 1)
sum += C[l / 2 + (d - 1) / 2];
}
return sum;
}
int main(){
int n, t, w;
while(scanf("%d%d%d", &n, &t, &w) == 3){
ll ans = 0;
init(t);
for(int i = 0; i < n; ++i){
int x, c;
scanf("%d%d", &x, &c);
ans += c * solve(x, w, t + 1);
ans %= mod;
}
printf("%lld\n", ans);
}
}