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hdu 5015 233 Matrix(最快的搞法)

2024-07-22 13:39:44   220人阅读

233 Matrix

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 191    Accepted Submission(s): 125


Problem Description
In our daily life we often use 233 to express our feelings. Actually, we may say 2333, 23333, or 233333 ... in the same meaning. And here is the question: Suppose we have a matrix called 233 matrix. In the first line, it would be 233, 2333, 23333... (it means a0,1 = 233,a0,2 = 2333,a0,3 = 23333...) Besides, in 233 matrix, we got ai,j = ai-1,j +ai,j-1( i,j ≠ 0). Now you have known a1,0,a2,0,...,an,0, could you tell me an,m in the 233 matrix?
 

Input
There are multiple test cases. Please process till EOF.

For each case, the first line contains two postive integers n,m(n ≤ 10,m ≤ 109). The second line contains n integers, a1,0,a2,0,...,an,0(0 ≤ ai,0 < 231).
 

Output
For each case, output an,m mod 10000007.
 

Sample Input
1 1
1
2 2
0 0
3 7
23 47 16
 

Sample Output
234
2799
72937


Hint
 

Source
2014 ACM/ICPC Asia Regional Xi‘an Online
 

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解释在代码里:


#include<cstdio>
#define MOD 10000007
using namespace std;
typedef long long LL;
#define rep(i,a,b) for(int i=a;i<b;i++)
LL inv[11]; ///逆元
LL inv_fac[11]; ///阶乘逆元
LL n,m,a[11],tmp;
/**
经过模拟之后发现ans=c(m,0)*a[n-1]+c(m+1,1)*a[n-2]+...c(m+n,n)*a[0]+c(n,0)*(23333...)+
c(n+1,1)*(2333...)+c(n+m-2,m-1)*233;
变形如下:
ans=c(m,0)*a[n-1]+c(m+1,1)*a[n-2]+...c(m+n,n)*a[0]+c(n,n)*(23333...)+
c(n+1,n)*(2333...)+c(n+m-2,n)*233;

ans=c(m,0)*a[n-1]+c(m+1,1)*a[n-2]+...c(m+n,n)*a[0]+{c(n,n)+
c(n+1,n)+c(n+m-2,n)}*233+2100*{10^(m-2)*c(n,n)+10^(m-3)*c(n+1,n)+10^0*c(n+m-4,n)};

ans=c(m,0)*a[n-1]+c(m+1,1)*a[n-2]+...c(m+n,n)*a[0]+c(n+m-1,n)*233
+2100*{10^(m-2)*c(n,n)+10^(m-3)*c(n+1,n)+10^0*c(n+m-4,n)};
设f(n)={10^(m-2)*c(n,n)+10^(m-3)*c(n+1,n)+10^0*c(n+m-4,n)},则有f(n)=((10*f(n-1)-c(n+m-2,n))/9)%MOD;
即f(n)=(10*f(n-1)-c(n+m-2,n)*inv[9]%MOD;
**/
void Init()///初始化数组
{
    int i;
    inv[0]=inv[1]=1;
    inv_fac[0]=inv_fac[1]=1;
    for(i=2;i<15;i++){
            inv[i]=(inv[MOD%i]*(MOD-MOD/i))%MOD;///递推求逆元
            inv_fac[i]=(inv_fac[i-1]*inv[i])%MOD;///递推求阶乘的逆元
    }
}

LL C(LL N,LL k)///计算组合数
{
    if(k==0)return 1;
    LL i,ans=inv_fac[k];
    for(i=0;i<k;i++)
        ans=ans*(N-i)%MOD;
    return ans;
}

LL quick_mod(LL x,LL N)///快速幂取模
{
    LL temp=x%MOD,result=1;
    while(N)
    {
        if(N&1)
            result=result*temp%MOD;
        temp=temp*temp%MOD;
        N>>=1;
    }
    return result;
}

LL f(LL N)
{
    if(N==0)
        return (tmp-1+MOD)*inv[9]%MOD;///用到了等比数列求和sum[n]=a1(1-q^n)/(1-q);
    return (10*f(N-1)-C(N+m-2,N))*inv[9]%MOD;
}
void solve()
{
    if(n==0&&m==0)
    {
        puts("0");
        return;
    }
    tmp=quick_mod(10,m-1);

    LL ans=(C(n+m-1,n)*233+f(n)*2100)%MOD;///第一行的和乘组合数
    int i;
    for(i=1;i<=n;i++)
        ans=(ans+C(n+m-1-i,n-i)*a[i-1])%MOD;///加上第一列的和乘组合数
    printf("%I64d\n",ans);
}
int main()
{
    Init();
    while(~scanf("%I64d%I64d",&n,&m))
    {
        rep(i,0,n){
            scanf("%I64d",&a[i]);
            a[i]%=MOD;
        }
        solve();
    }
    return 0;
}

hdu 5015 233 Matrix(最快的搞法)


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