bzoj4563 [Haoi2016]放棋子

Description

---

给你一个N*N的矩阵，每行有一个障碍，数据保证任意两个障碍不在同一行，任意两个障碍不在同一列，要求你在
这个矩阵上放N枚棋子（障碍的位置不能放棋子），要求你放N个棋子也满足每行只有一枚棋子，每列只有一枚棋子
的限制，求有多少种方案。

第一行一个N，接下来一个N*N的矩阵。N<=200,0表示没有障碍，1表示有障碍，输入格式参考样例

Solution

---

首先可以想到行与行之间、列与列之间可以交换而不改变答案。把障碍移到对角线上后可以发现，一个合法的答案当且仅当第i行的棋子不在第i列，且每一行都有棋子。这是一个错排的模型。没有模数上高精度

Code

---

#include <stdio.h>
#include <string.h>
#include <algorithm>
#define rep(i,st,ed) for (int i=st;i<=ed;++i)
#define drp(i,st,ed) for (int i=st;i>=ed;--i)
#define fill(x,t) memset(x,t,sizeof(x))

const int MOD=1000;
const int N=505;
const int L=505;

struct num {
    int s[L], len;
    inline bool operator ==(num b) {
        num a = *this;
        if (a.len != b.len) {
            return 0;
        }else {
            drp(i, a.len, 1) {
                if (a.s[i] != b.s[i]) {
                    return 0;
                }
            }
            return 1;
        }
    }
    inline bool operator <(num b) {
        num a = *this;
        if (a.len < b.len) {
            return 1;
        }else if (a.len > b.len) {
            return 0;
        }else {
            drp(i, a.len, 1) {
                if (a.s[i] < b.s[i]) {
                    return 1;
                }else if (a.s[i] > b.s[i]) {
                    return 0;
                }
            }
            return 0;
        }
    }
    inline bool operator <=(num b) {
        num a = *this;
        if (a < b || a == b) {
            return 1;
        }else {
            return 0;
        }
    }
    inline bool operator >(num b) {
        num a = *this;
        if (a.len > b.len) {
            return 1;
        }else if (a.len < b.len) {
            return 0;
        }else {
            drp(i, a.len, 1) {
                if (a.s[i] > b.s[i]) {
                    return 1;
                }else if (a.s[i] < b.s[i]) {
                    return 0;
                }
            }
            return 0;
        }
    }
    inline bool operator >=(num b) {
        num a = *this;
        if (a > b || a == b) {
            return 1;
        }else {
            return 0;
        }
    }
    inline num operator +(num b) {
        num a = *this, c = (num) { {0}, std:: max(a.len, b.len)};
        int v = 0;
        rep(i, 1, c.len) {
            c.s[i] = (a.s[i] + b.s[i] + v) % MOD;
            v = (a.s[i] + b.s[i] + v) / MOD;
        }
        if (v) {
            c.len += 1;
            c.s[c.len] = v;
        }
        return c;
    }
    inline num operator -(num b) {
        num a = *this, c = (num) { {0}, std:: max(a.len, b.len)};
        rep(i, 1, c.len) {
            c.s[i] = a.s[i] - b.s[i];
            if (c.s[i] < 0) {
                c.s[i] += MOD;
                a.s[i + 1] -= 1;
            }
        }
        while (!c.s[c.len] && c.len > 1) {
            c.len -= 1;
        }
        return c;
    }
    inline num operator *(num b) {
        num a = *this, c = (num) { {0}, a.len + b.len};
        rep(i, 1, a.len) {
            rep(j, 1, b.len) {
                c.s[i + j - 1] += a.s[i] * b.s[j];
            }
        }
        rep(i, 1, a.len + b.len) {
            c.s[i + 1] += c.s[i] / MOD;
            c.s[i] %= MOD;
        }
        while (!c.s[c.len] && c.len > 1) {
            c.len -= 1;
        }
        return c;
    }
    inline num operator /(int b) {
        num a = *this, c = (num) { {0}, a.len};
        int v = 0;
        drp(i, len, 1) {
            int t = v * MOD + a.s[i];
            c.s[i] = t / b;
            v = t % b;
        }
        while (!c.s[c.len] && c.len > 1) {
            c.len -= 1;
        }
        return c;
    }
    inline void read() {
        fill(s, 0); len = 0;
        char st[L];
        scanf("%s", st);
        int v = 0, i;
        for (i = strlen(st) - 1; i >= 3; i -= 3) {
            rep(j, i - 2, i) {
                v = v * 10 + st[j] - '0';
            }
            s[++ len] = v;
            v = 0;
        }
        rep(j, 0, i) {
            v = v * 10 + st[j] - '0';
        }
        s[++ len] = v;
    }
    inline void read1(int x) {
        fill(s, 0);
        len = 0;
        do {s[++ len] = x % MOD;}while (x /= MOD);
    }
    inline void output() {
        num tmp = *this;
        int i = tmp.len;
        while (!tmp.s[i] && i > 1) {
            i -= 1;
        }
        printf("%d", tmp.s[i]);
        drp(j, i - 1, 1) {
            int v = tmp.s[j];
            int f[5] =  {0};
            rep(k, 1, 3) {
                f[k] = v % 10;
                v /= 10;
            }
            drp(k, 3, 1) {
                printf("%d", f[k]);
            }
        }
        printf("\n");
    }
} f[N];
num one;

int main(void)  {
    int n; scanf("%d",&n);
    rep(i,1,n) rep(j,1,n) scanf("%*d");
    f[2].len=1; f[2].s[1]=1;
    rep(i,3,n) {
        num tmp; tmp.read1(i-1);
        f[i]=f[i-1]+f[i-2];
        f[i]=f[i]*tmp;
    }
    f[n].output();
    return 0;
}