2019ACM-ICPC沈阳网络赛-K-Guanguan's Happy water

Guanguan's Happy water

4000ms 

262144K

 

Rather than drinking happy water, Guanguan loves storing happy water. So he bought a refrigerator and stored a_iai​ bottles of cola into it every day. When the storage is finished on the kk-th day, the refrigerator is full, but he still wants to store happy water every day. Here comes the solution: He first constructs a p-sequence: p_1p1​, p_2p2​, ..., p_kpk​, where p_1+p_2+...+p_k=1p1​+p2​+...+pk​=1. Then he chooses an number ii among 11 to kk, where number ii has the probability p_ipi​ to be chosen. After that, he drinks the happy water stored on the ii-th day before the current day and stores the same amount of happy water back into the refrigerator again. Let the amount of happy water stored on the ii-th day be f_ifi​. Given the amount of happy water stored in the first kk days and the expected amount of the next kk days(which means, from the k+1k+1-th day to the 2k2k-th day), could you help Guanguan figure out the sum of the expected amount of happy water stored during the first nn days) (Be aware that every element of ff has moded 1e9+71e9+7 when input datas, and your output should mod 1e9+71e9+7 as well)

Input

The first line is TT (1 le T le 201≤T≤20), indicating the number of input sets. For each set of inputs, the first line is kk and nn (1 le k le 701≤k≤70, 1 le n le 10^{18}1≤n≤1018), and the second line is 2k2k numbers, respectively representing a_1a1​, a_2a2​, ..., a_kak​, f_{k+1}fk+1​, f_{k+2}fk+2​, ..., f_{2k}f2k​.

Output

For each data, output a non-negative integer indicating (sum_{i=1}^n f_i) mod 10^9+7(∑i=1n​fi​)mod109+7.

样例输入复制

2
1 9
2 2
2 8
6 5 5 5

样例输出复制

18
41

思路：

suma=(a1+a2+。。。+ak)

sumf=(f1+f2+。。。+f3)

①n<=k,ans=a1+a2+。。。+an;

②n<=2k,ans=suma+f1+f2+。。。+f(n-k);

③n>2k,ans=suma+((n-k)/k)*sumf+(   f1+f2+。。。+f( (n-k)%k )  ).

注意取模就行

AC代码：

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=1e6+7;
ll mod=1e9+7;
ll a[maxn],f[maxn];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--){
        int k;
        ll n,suma=0,sumf=0,k2;
        scanf("%d",&k);
        k2=ll(k);
        scanf("%lld",&n);
        for(int i=1;i<=k;++i)
            scanf("%lld",&a[i]),suma=(suma+a[i])%mod;
        for(int i=1;i<=k;++i)
            scanf("%lld",&f[i]),sumf=(sumf+f[i])%mod;
        ll ans=0;
        if(k2>=n){
            for(int i=1;i<=int(n);++i)
                ans=(ans+a[i])%mod;
            printf("%lld
",ans%mod);
        }
        else{
            ans=suma%mod;
            ll nu=((n-k2)/k2);
            int len=n-nu*k2-k2;
            nu%=mod;
            ans=(nu*sumf+ans)%mod;
            for(int i=1;i<=len;++i)
                ans=(ans+f[i])%mod;
            printf("%lld
",ans%mod);
        }
    }
    return 0;
}
/*
2
2 100000000000000000
6 5 5 5
*/