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fireworks

Time Limit: 1000MS  Memory Limit: 65536KB Submit  Statistic

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

Hint

也就是组合数取模

#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9+7;
const int maxn = 1e5+10;
typedef long long LL;
LL fac[maxn];
LL modxp(LL a,LL b){
    LL res = 1;
    while(b){
        if(b&1)
            res = (res*a)%mod;
        b = b>>1;
        a = (a*a)%mod;
    }
    return res;
}
void Init(){
    fac[0] = 1;
    for(int i = 1; i < maxn; i++){
        fac[i] = (fac[i-1]*i)%mod;
    }
}
LL C(LL a,LL k){
    LL re = 1;
    while(a && k){
        LL aa = a%mod;
        LL bb = k%mod;
        if(aa<bb)
            return 0;
        re = re*fac[aa]*modxp(fac[bb]*fac[aa-bb]%mod,mod-2)%mod;
        a /= mod;
        k /= mod;
    }
    return re;
}
int main(){
    LL n,t,w,c,x,ans;
    Init();
    while(~scanf("%lld %lld %lld",&n,&t,&w)){
        ans = 0;
        while(n--){
            scanf("%lld %lld",&x,&c);
            LL l = x-t;
            LL r = x+t;
            LL temp = (w-l);
            if(w<l || w>r || temp%2!=0)
                continue;
            ans = (c*C(t,temp/2))%mod+ans;
            ans = ans%mod;
        }
        printf("%lld\n",ans);
    }
    return 0;
}

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