Lightoj 1132 Summing up Powers(矩阵快速幂)

题意：给你n和k，求(1^k + 2^k + 3^k + ... + n^k) % 2^32， n <= 1e15, k <= 50

思路：可以知道f(x+1) = f(x) + (x+1)^k,

(x+1)^k 可以用二项式定理展开变成C(k, 0)*x^k+C(k, 1)*x^(k-1)...+C(k, k)*x^0

这就可以构造矩阵快速幂了。

具体思路：

代码：

#include<iostream>#include<cstdio>#include<cstring>using namespace std;typedef long long ll;const int maxn = 55;unsigned int c[maxn][maxn];ll n, K;struct node{    unsigned int s[maxn][maxn];};node mul(node a, node b){    node t;    memset(t.s, 0, sizeof(t.s));    for(int i = 0; i < K+2; i++)        for(int k = 0; k < K+2; k++)            for(int j = 0; j < K+2; j++)                t.s[i][j] = t.s[i][j]+a.s[i][k]*b.s[k][j];    return t;}node mt_pow(node p, ll k){    node q;    memset(q.s, 0, sizeof(q.s));    for(int i = 0; i < K+2; i++)        q.s[i][i] = 1;    while(k)    {        if(k%2) q = mul(p, q);        p = mul(p, p);        k /= 2;    }    return q;}void init(){    for(int i = 1; i < maxn; i++)        for(int j = 0; j <= i; j++)        {            if(!j || j==i) c[i][j] = 1;            else c[i][j] = c[i-1][j]+c[i-1][j-1];         }}int main(void){    init();//    freopen("out.txt", "w", stdout);    int _, ca = 1;    cin >> _;    while(_--)    {        scanf("%lld%lld", &n, &K);        if(K == 0) printf("Case %d: %u\n", ca++, (unsigned int)n);        else        {            node base;            memset(base.s, 0, sizeof(base.s));            base.s[0][0] = 1;            for(int i = 1; i < K+2; i++) base.s[0][i] = c[K][i-1];            for(int i = 1; i < K+1; i++)                for(int j = i; j < K+2; j++)                    base.s[i][j] = c[K-i+1][j-i];            base.s[K+1][K+1] = 1;            node ans = mt_pow(base, n-1);            unsigned int res = 0;            for(int i = 0; i < K+2; i++)                res += ans.s[0][i];            printf("Case %d: %u\n", ca++, res);        }    }    return 0;}