Itself is Itself（scu1617）

1617: Itself is Itself

Time Limit: 6 Sec   Memory Limit: 128 MB
Submit: 81   Solved: 18
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Description

Zuosige always has bad luck. Recently, he is in hospital because of pneumonia. While he is taking his injection, he feels extremely bored. However, clever Zuosige comes up with a new game.

Zuosige writes an integer n and a polynomial function:
P(x) = (a₀+a₁*x+a₂*x²+…+a_m*x^m) mod n.
He wants to know how many subsets of set {0, 1, 2, … , n-2, n-1} satisfies following property: the range of P(x) from the subset is itself. Pay attention to that empty set is also a valid subset.

Input

The first line contains one integer T, indicating the number of test cases.
In one test case, there are two lines.
In the first line, there are two integers n and m (1<=n<=10000, 1<=m<=1000).
In the second line, there m+1 integers. The i-th integer indicating a_i-1 (0<=a_i<=10000).

Output

For each test case, output an integer in a line indicating the answer. The answer can be very large, so you need just output the answer mod 1e9+7.

Sample Input

2
3 1
0 1
3 1
1 1

Sample Output

8
2

HINT

Source

 

图论（强连通）数圈数n，2^n%mod=

 

可惜比赛的时候没过。。。

 

转载请注明出处：http://blog.csdn.net/u010579068

 

题目链接：http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1617

 

#include<stdio.h>
#include<vector>
#include<string.h>
using namespace std;
#define modd 1000000007
const int N= 10005;
 
int n,c[N];
int dfn[N],low[N],Stack[N],flag[N],vist[N],num[N],top,deep,tn;
vector<int>mapt1[N];
 
void init()
{
    for(int i=0;i<=n;i++)
    {
        mapt1[i].clear();
        dfn[i]=0;
        num[i]=0;
        vist[i]=0;
    }
    tn=top=deep=0;
}
int tt;
void tarjan(int u)
{
    vist[u]=tt;
    deep++;
    dfn[u]=low[u]=deep;
    Stack[++top]=u;
 
    int len=mapt1[u].size();
    for(int i=0;i<len;i++)
    {
        int v=mapt1[u][i];
 
        if(vist[v]==0)
        {
            tarjan(v);
            if(low[u]>low[v])
                low[u]=low[v];
        }
        else if(vist[v]==tt&&low[u]>dfn[v])//注意vist[v]==tt
            low[u]=dfn[v];
    }
    if(low[u]==dfn[u])
    {
        tn++;
        while(u!=Stack[top])
        {
            flag[Stack[top]]=tn; num[tn]++;top--; //printf("%d ",Stack[top]);
        }
        flag[Stack[top]]=tn; num[tn]++; top--;
    }
}
int out[N];
int rebuilMap()//用强连通缩点，重新反向建图
{
    tt=0;
    for(int i=0;i<n;i++)
        if(vist[i]==0)
        {
            tt++;
            tarjan(i);
        }
 
    memset(out,0,sizeof(out));
//缩点后的图无 有向环
    for(int i=0;i<n;i++)
    {
        int u=flag[i];
        for(int j=0;j<mapt1[i].size();j++)
        {
            int v=flag[mapt1[i][j]];
            if(u==v)
                continue;
            out[u]++;
        }
    }
    int number=0;
    for(int i=1;i<=tn;i++)
        if(out[i]==0)
            number++;
    return number;
}
int main()
{
    int T,m;
 
    scanf("%d",&T);
    while(T--){
        scanf("%d%d",&n,&m);
        for(int i=0;i<=m;i++){
            scanf("%d",&c[i]);
            c[i]%=n;
        }
 
        init();
        for(int i=0;i<n;i++)
        {
            int ansa=c[0],x=i;
            for(int j=1;j<=m;j++){
                    ansa=(ansa+c[j]*x)%n;
                    x=(x*i)%n;
            }
            if(i!=ansa)
            mapt1[i].push_back(ansa);
        }
 
        int number=rebuilMap();
        long long ans = 1;
        for(int i=1;i<=number;i++)
            ans=(ans*2)%1000000007;
        printf("%lld\n",ans);
    }
    return 0;
}
 
/**************************************************************
    Problem: 1617
    User: aking2015
    Language: C++
    Result: Accepted
    Time:2756 ms
    Memory:1628 kb
****************************************************************/