题目连接：uva 766 - Sum of powers

题目大意：将Sk(n)=∑i=1nik化简成Sk(n)=ak+1nk+1+aknk+?+a0M

解题思路：

已知幂k，并且有(n+1)k=C(kk)nk+C(k?1k)nk?1+?+C(0k)n0结论。
所以令 (n+1)k+1?nk+1=C(kk+1)nk+C(k?1k+1)nk?1+?+C(0k+1)n0
nk+1?(n?1)k+1=C(kk+1)(n?1)k+C(k?1k+1)(n?1)k?1+?+C(0k+1)(n?1)0
…
2k+1?1k+1=C(kk+1)1k+C(k?1k+1)1k?1+?+C(0k+1)10
将各项累加起来的(n+1)k?1=C(kk+1)Sk(n)+C(k?1k+1)Sk?1(n)+?+C(0k+1)S0(n)

#include <cstdio>
#include <cstring>

typedef long long ll;
const int N = 25;


ll m[N], a[N][N];
ll C[N][N];

ll gcd (ll a, ll b) {
    return b == 0 ? a : gcd(b, a%b);
}

inline ll cal (ll d) {
    if (d < 0)
        return -d;
    return d;
}

inline ll sign (ll d) {
    if (d > 0)
        return 1;
    else if (d < 0)
        return -1;
    else
        return 0;
}

void del (ll& t, ll* p, int n) {
    ll d = gcd(t, p[0]);
    for (int i = 1; i <= n; i++) {
        if (p[i] == 0)
            continue;
        d = gcd(d, cal(p[i]));
    }

    t /= d;
    for (int i = 0; i <= n; i++)
        p[i] = cal(p[i]) / d * sign(p[i]);
}

void add (ll* p, ll* q, ll k, ll& t, ll f, int n) {

    for (int i = 0; i <= n; i++)
        p[i] = p[i] * f - q[i] * k * t;
    t *= f;

    del(t, p, n);
}

void init () {
    C[0][0] = 1;
    for (int i = 1; i < N; i++) {
        C[0][i] = C[i][i] = 1;
        for (int j = 1; j < i; j++)
            C[j][i] = C[j-1][i-1] + C[j][i-1];
    }

    memset(a, 0, sizeof(a));
    m[0] = 1;
    a[0][1] = 1;

    for (int i = 1; i <= 20; i++) {

        int u = i+1;
        for (int j = 1; j <= u; j++)
            a[i][j] += C[j][u];

        m[i] = 1;
        for (int j = 0; j < i; j++) {
            add (a[i], a[j], C[j][u], m[i], m[j], u);

            /*
            for (int x = 0; x <= j + 1; x++) {
                a[i][x] -= (a[j][x] * C[j][u] * tmp / m[j]);
            }
            */
        }

        m[i] *= C[i][u];
        del(m[i], a[i], u);
    }
}

int main () {
    init();
    int cas, k;
    scanf("%d", &cas);

    while (cas--) {
        scanf("%d", &k);
        printf("%lld", m[k]);
        for (int i = k+1; i >= 0; i--)
            printf(" %lld", a[k][i]);
        printf("\n");
        if (cas)
            printf("\n");
    }
    return 0;
}