[bzoj5339][TJOI2018]教科書般的褻瀆【拉格朗日插值法】
阿新 • • 發佈:2019-01-11
# include <bits/stdc++.h> # define ll long long # define inf 0x3f3f3f3f # define N 110 using namespace std; int read(){ int tmp = 0, fh = 1; char ch = getchar(); while (ch < '0' || ch > '9') {if (ch == '-') fh = -1; ch = getchar(); } while (ch >= '0' && ch <= '9'){tmp = tmp * 10 + ch - '0'; ch = getchar(); } return tmp * fh; } ll readll(){ ll tmp = 0, fh = 1; char ch = getchar(); while (ch < '0' || ch > '9') {if (ch == '-') fh = -1; ch = getchar(); } while (ch >= '0' && ch <= '9'){tmp = tmp * 10 + ch - '0'; ch = getchar(); } return tmp * fh; } const int P = 1e9 + 7; ll num[N], n; int m, x[N], y[N]; int power(int x, int y){ int i = x; x = 1; while (y > 0){ if (y % 2 == 1) x = 1ll * x * i % P; i = 1ll * i * i % P; y /= 2; } return x; } int getsum(ll tmp, int lim){ int sum = 0; tmp %= P; for (int i = 1; i <= lim; i++){ int num1 = 1, num2 = 1; for (int j = 1; j <= lim; j++) if (i != j){ num1 = 1ll * num1 * (tmp - x[j]) % P; num2 = 1ll * num2 * (x[i] - x[j]) % P; } sum = (sum + 1ll * num1 * y[i] % P * power(num2, P - 2)) % P; } return sum; } int main(){ // freopen(".in", "r", stdin); // freopen(".out", "w", stdout); for (int opt = read(); opt--;){ n = readll(), m = read(); for (int i = 1; i <= m; i++) num[i] = readll(); sort(num + 1, num + m + 1); int k = m + 1; for (int i = 1; i <= k + 2; i++) x[i] = i, y[i] = (y[i - 1] + power(i, k)) % P; int ans = 0; for (int i = 1; i <= k; i++){ ll now = n - num[i - 1]; ans = (ans + getsum(now, k + 2)) % P; for (int j = i; j <= m; j++) ans = (ans - power(num[j] - num[i - 1], k)) % P; } printf("%d\n", (ans + P) % P); } return 0; }