牛客寒假训练营2-C算概率

思路

　　用 f(i,j) 来表示当前做了i道题，共做对了j道题

　　状态 f[i][j] = f[i-1][j] * (1-p[i]) + f[i-1][j-1] * p[i]

　　第一种：由于i-1时对了j题，所以第i题做错了；

　　第二种：由于i-1时对了j-1题，所以第i题对了；

　　时间复杂度O（n^2)

CODE

#include <bits/stdc++.h>
#define dbg(x) cout << #x << "=" << x << endl

using namespace std;
typedef long long LL;

template<class T>inline void read(T &res)
{
    char c;T flag=1;
    while((c=getchar())<'0'||c>'9')if(c=='-')flag=-1;res=c-'0';
    while((c=getchar())>='0'&&c<='9')res=res*10+c-'0';res*=flag;
}

namespace _buff {
    const size_t BUFF = 1 << 19;
    char ibuf[BUFF], *ib = ibuf, *ie = ibuf;
    char getc() {
        if (ib == ie) {
            ib = ibuf;
            ie = ibuf + fread(ibuf, 1, BUFF, stdin);
        }
        return ib == ie ? -1 : *ib++;
    }
}

int qread() {
    using namespace _buff;
    int ret = 0;
    bool pos = true;
    char c = getc();
    for (; (c < '0' || c > '9') && c != '-'; c = getc()) {
        assert(~c);
    }
    if (c == '-') {
        pos = false;
        c = getc();
    }
    for (; c >= '0' && c <= '9'; c = getc()) {
        ret = (ret << 3) + (ret << 1) + (c ^ 48);
    }
    return pos ? ret : -ret;
}

const int kmaxn = 1007;
LL f[kmaxn][kmaxn];
const LL mod = 1e9 + 7;
int n;
LL p[kmaxn];

int main()
{
    scanf("%d",&n);
    for(int i = 1; i <= n; ++i) {
        scanf("%lld",&p[i]);
    }
    f[0][0] = 1;
    for(int i = 1; i <= n; ++i) {
        f[i][0] = f[i-1][0] * (mod+1-p[i]) % mod;
        for(int j = 1; j <= n; ++j) {
            f[i][j] = ((f[i-1][j]*(i-p[i]+mod)) % mod + f[i-1][j-1]*(mod+p[i]) % mod)% mod;
        }
    }
    for(int i = 0; i < n; ++i) {
        printf("%lld ",f[n][i]);
    }
    printf("%lld
",f[n][n]);
    return 0;
}