BZOJ1856: [Scoi2010]字元串(組合數)

題意

題目鏈接

Sol

\(30 \%\)dp：

\(f[i][j]\)表示放了\(i\)個\(1\)和\(j\)個\(0\)的不合法方案

    f[0][0] = 1;
    cin >> N >> M;
    for(int i = 1; i <= N; i++) {
        f[i][0] = 1;
        for(int j = 1; j <= i; j++) {
            f[i][j] = add(f[i - 1][j], f[i][j - 1]);
        }
    }
    cout << f[N][M];

我們可以把\(1\)看做是\((+1, +1)\), \(0\)看做是\((+1, -1)\)，根據折射原理，不合法的方案為\(C_{n+m}^{n+1}\)

詳細點的題解可以看這裡

#include<bits/stdc++.h>
#include<algorithm>
#define LL long long 
#define ull long long 
using namespace std;
const int MAXN = 2e6 + 10, mod = 20100403;
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, fac[MAXN], ifac[MAXN];
int C(int N, int M) {
    return mul(mul(fac[N], ifac[M]), ifac[N - M]);
}
main() {
    cin >> N >> M; int Lim = N + M;
    fac[0] = 1;
    for(int i = 1; i <= Lim; i++) fac[i] = mul(i, fac[i - 1]);
    ifac[Lim] = fp(fac[Lim], mod - 2);
    for(int i = Lim; i >= 1; i--) ifac[i - 1] = mul(ifac[i], i);
    printf("%d\n", (C(N + M, N) - C(N + M, N + 1) + mod) % mod);
    return 0;
}