BZOJ3907

来源：互联网发布：windows找不到runtime 编辑：程序博客网时间：2024/05/23 15:57

题解是粘的代码也是粘的

转自这里

(0,0)->(n,m)方案数为C(n,n+m), 然后减去不合法的方案. 作(n,m)关于y=x+1的对称点(m-1,n+1), 则(0,0)->(m-1,n+1)的任意一条路径都对应(0,0)->(n,m)的一条不合法路径(y>x). 所以答案就是C(n,n+m) - C(n+1,n+m).高精度算就OK了

#include<cstring>#include<algorithm>using namespace std;const int maxn = 10009;const int maxpn = 1500;int N, M;int p[maxpn], pn = 0;int a[2][maxpn];bool check[maxn];void init() {    scanf("%d%d", &N, &M);    memset(check, 0, sizeof check);    for(int i = 2, lim = N + M; i <= lim; i++) {        if(!check[i]) p[pn++] = i;        for(int j = 0; j < pn && i * p[j] <= lim; j++) {            check[i * p[j]] = true;            if(i % p[j] == 0) break;        }    }    memset(a, 0, sizeof a);    for(int i = N + 2; i <= N + M; i++)        for(int j = 0, t = i; j < pn && t >= p[j]; j++)            for(; t % p[j] == 0; a[0][j]++, a[1][j]++, t /= p[j]);    for(int t = N + 1, j = 0; j < pn && t >= p[j]; j++)        for(; t % p[j] == 0; a[0][j]++, t /= p[j]);     for(int i = 2; i < M; i++)        for(int j = 0, t = i; j < pn && t >= p[j]; j++)            for(; t % p[j] == 0; a[0][j]--, a[1][j]--, t /= p[j]);    for(int t = M, j = 0; j < pn && t >= p[j]; j++)        for(; t % p[j] == 0; a[0][j]--, t /= p[j]);}struct Int {    static const int base = 10000;    static const int maxn = 1000;    static const int width = 4;    int s[maxn], n;    Int() : n(0) {        memset(s, 0, sizeof s);    }    Int operator = (int num) {        n = 0;        for(; num; num /= base) s[n++] = num % base;        return *this;    }    Int(int x) {        *this = x;    }    Int operator = (const Int &o) {        n = o.n;        memcpy(s, o.s, sizeof(int) * n);        return *this;    }    void write() {        int buf[width];        for(int i = n; i--; ) {            int t = s[i], c = 0;            for(; t; t /= 10) buf[c++] = t % 10;            if(i + 1 != n)                for(int j = width - c; j; j--) putchar('0');            while(c--) putchar(buf[c] + '0');        }        puts("");    }    Int operator + (const Int &o) const {        Int res;        for(int i = 0, g = 0; ; i++) {            if(!g && i >= o.n && i >= n) break;            int x = g + (i < n ? s[n] : 0) + (i < o.n ? o.s[o.n] : 0);            res.s[res.n++] += x / base;            g = x % base;        }        return res;    }    Int operator += (const Int &o) {        return (*this = *this + o);    }    Int operator * (const Int &o) const {        Int ret; ret.n = o.n + n - 1;        for(int i = 0; i < n; i++)            for(int j = 0; j < o.n; j++) {                int t = s[i] * o.s[j];                ret.s[i + j] += t % base;                ret.s[i + j + 1] += t / base;            }        for(int i = 0; i < ret.n; i++) if(ret.s[i] >= base) {            ret.s[i + 1] += ret.s[i] / base;            ret.s[i] %= base;        }        for(int &i = ret.n; ret.s[i]; i++) {            ret.s[i + 1] += ret.s[i] / base;            ret.s[i] %= base;        }        return ret;    }    Int operator *= (const Int &o) {        return (*this = *this * o);    }    Int operator - (const Int &o) const {        Int ret = *this;        for(int i = ret.n; --i; ) {            ret.s[i]--; ret.s[i - 1] += base;        }        for(int i = 0; i < o.n; i++)            if((ret.s[i] -= o.s[i]) >= base)                ret.s[i + 1]++, ret.s[i] -= base;        if(!ret.s[n - 1]) ret.n--;        return ret;    }};Int power(int p, int a) {    Int ret = 1, _p = p;    for(; a; _p *= _p, a >>= 1) if(a & 1) ret *= _p;    return ret;}int main() {    init();    Int ans = 1, t = 1;    for(int i = 0; i < pn; i++)        ans *= power(p[i], a[0][i]), t *= power(p[i], a[1][i]);    (ans - t).write();    return 0;}

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