BZOJ4559: [JLoi2016]成績比較(dp 拉格朗日插值)
阿新 • • 發佈:2018-12-01
題意
Sol
想不到想不到。。
首先在不考慮每個人的真是成績的情況下,設\(f[i][j]\)表示考慮了前\(i\)個人,有\(j\)個人被碾壓的方案數
轉移方程:\[f[i][j] = \sum_{k = j}^n f[i -1][k] C_{k}^{k - j} C_{N - k}^{r[i] - 1 - (k - j)} * g(i)\]
大概解釋一下,列舉的\(k\)表示之前碾壓了多少,首先我們湊出\(j\)個人繼續碾壓,也就是說會有\(k - j\)個人該課的分數比\(B\)爺高,那麼這\(k\)個人我們已經考慮完了
接下來需要從剩下的\(N-k\)個人中,選出\(r[i] - 1 - (k - j)\)
後面的\(g(i)\)表示的是吧\(1 \sim U_i\)的分數,分給\(N\)個人後,有\(R_i\)個人比B爺高的方案數
這個計算的時候可以直接列舉B爺的分數
\(g(k) = \sum_{i = 1}^{U_k} i^{N - R[i]} * (U_k - i) ^{R[i] - 1}\)
後面的次數小於等於\(N-1\),然後直接插值一下就行了
// luogu-judger-enable-o2 #include<bits/stdc++.h> using namespace std; const int MAXN = 103, mod = 1e9 + 7; inline int read() { char c = getchar(); int x = 0, f = 1; while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();} while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar(); return x * f; } int add(int x, int y) { if(x + y < 0) return x + y + mod; return x + y >= mod ? x + y - mod : x + y; } int mul(int x, int y) { return 1ll * x * y % mod; } int fp(int a, int p) { int base = 1; while(p) { if(p & 1) base = mul(base, a); a = mul(a, a); p >>= 1; } return base; } int N, M, K, f[MAXN][MAXN], C[MAXN][MAXN], U[MAXN], R[MAXN], g[MAXN]; int get(int U, int R) { memset(g, 0, sizeof(g)); for(int i = 1; i <= MAXN - 1; i++) for(int k = 1; k <= i; k++) g[i] = add(g[i], mul(fp(k, N - R), fp(i - k, R - 1))); int ans = 0; for(int i = 1; i <= MAXN - 1; i++) { int up = 1, down = 1; for(int j = 1; j <= MAXN - 1; j++) { if(i == j) continue; up = mul(up, add(U, -j)); down = mul(down, add(i, -j)); } ans = add(ans, mul(g[i], mul(up, fp(down, mod - 2)))); } return ans; } int main() { //freopen("a.in", "r", stdin); N = read(); M = read(); K = read(); for(int i = 0; i <= N; i++) { C[i][0] = C[i][i] = 1; for(int j = 1; j < i; j++) C[i][j] = add(C[i - 1][j - 1], C[i - 1][j]); } for(int i = 1; i <= M; i++) U[i] = read(); for(int i = 1; i <= M; i++) R[i] = read(); f[0][N - 1] = 1; for(int i = 1; i <= M; i++) { int t = get(U[i], R[i]); for(int j = K; j <= N; j++) { for(int k = j; k <= N - 1; k++) if(k - j <= R[i] - 1) f[i][j] = add(f[i][j], mul(mul(f[i - 1][k], C[k][k - j]), C[N - 1 - k][R[i] - 1 - (k - j)])); f[i][j] = mul(f[i][j], t); } } printf("%d", f[M][K]); return 0; } /* 100 3 50 500 500 456 13 46 45 */