Light OJ 1132 Summing up Powers 矩阵快速幂

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题目大意：

给出 N,k， 1 <= N <= 10^15, 0 <= k <= 50;

求( 1^k + 2^k + ... + N^k) mod (2^32) 的值

大致思路：

是个很明显的矩阵快速幂的题吧，之前学长推荐过这道题，现在居然正好碰到了

不过刚开始一直因为 %lld 和 %I64d 的原因WA了3次..以后还是要注意OJ上的FAQ的问题...

设S(n) = sigma(i^k) mod (2^32) i 从1到n

这道题的关键就是发现这样的转换关系：

这样转换矩阵当中的数之和K有关，这样就好办了

直接用快速幂，最后得到的S(n)就是答案

代码如下：

Result : Accepted Memory : 1716 KB Time : 1724 ms

/* * Author: Gatevin * Created Time:  2014/7/29 18:29:03 * File Name: test.cpp */#include<iostream>#include<sstream>#include<fstream>#include<vector>#include<list>#include<deque>#include<queue>#include<stack>#include<map>#include<set>#include<bitset>#include<algorithm>#include<cstdio>#include<cstdlib>#include<cstring>#include<cctype>#include<cmath>#include<ctime>#include<iomanip>using namespace std;const double eps(1e-8);typedef long long lint;lint k,siz;lint n;const lint mod = ((long long)1 << 32);lint C[60][60];struct Matrix{    lint a[53][53];    Matrix()    {        memset(a, 0, sizeof(a));        for(int i = 1; i <= 52; i++)        {            a[i][i] = 1LL;        }    }};Matrix operator * (const Matrix & m1, const Matrix & m2){    Matrix m;    for(int i = 1; i <= siz; i++)    {        for(int j = 1; j <= siz; j++)        {            m.a[i][j] = 0;            for(int k = 1; k <= siz; k++)            {                m.a[i][j] = (m.a[i][j] + (m1.a[i][k] * m2.a[k][j]) % mod) % mod;            }        }    }    return m;}Matrix quick_pow(Matrix base, lint pow){    Matrix I;    while(pow)    {        if(pow & 1)        {            I = I * base;        }        base = base * base;        pow >>= 1;    }    return I;}int main(){    lint t;      memset(C, 0, sizeof(C));    C[0][0] = 1LL;    C[1][0] = C[1][1] = 1LL;    for(int i = 2; i <= 50; i++)    {        for(int j = 0; j <= i; j++)        {            if(j == 0)            {                C[i][j] = 1LL;            }            else            {                C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % mod;            }        }    }    scanf("%lld",&t);    for(lint cas = 1; cas <= t; cas++)    {        scanf("%lld %lld", &n, &k);        siz = k + 2;        Matrix tran;        memset(tran.a, 0, sizeof(tran.a));        for(int i = 1; i <= k + 1; i++)        {            for(int j = 1; j <= k + 1; j++)            {                if(i <= j)                {                    tran.a[i][j] = C[k + 1 - i][j - i];                }            }        }        for(int i = 1; i <= k + 1; i++)        {            tran.a[k + 2][i] = C[k][i - 1];        }        tran.a[k + 2][k + 2] = 1LL;        Matrix ans = quick_pow(tran, n - 1);        lint answer = 0;        for(int i = 1; i <= k + 2; i++)        {            answer = (answer + ans.a[k + 2][i]) % mod;        }        printf("Case %lld: %lld\n", cas, answer);    }    return 0;}

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