CodeForces451E Devu and Flowers

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problem analysis

I did not think the practice of female function ......

In fact, a direct look at the topic and ideas quite simple. If you find that each flower has an infinite number of words, the problem becomes very simple, the answer is \ (the n-S-1 + \ the Choose the n-- 1 \) . Then find \ (n \) only \ (20 \) , so vigorously to get away with a wave of inclusion and exclusion.

Reference Code

#include <cstdio>

const long long Max_n = 30;
const long long Mod = 1000000007;
long long n, s, f[ Max_n ];

void Exgcd( long long a, long long b, long long & x, long long & y ) {
    if( b == 0LL ) { x = 1LL; y = 0LL; return; }
    Exgcd( b, a % b, y, x );
    y -= a / b * x;
    return;
}

long long Inv( long long a ) {
    long long x, y;
    Exgcd( a, Mod, x, y );
    if( x < 0 ) x += Mod;
    return x;
}

long long C( long long n, long long m ) {
    long long Ans = 1;
    for( long long i = 1; i <= m; ++i ) Ans = Ans * ( ( n - i + 1 ) % Mod ) % Mod;
    for( long long i = 1; i <= m; ++i ) Ans = Ans * Inv( i ) % Mod;
    return Ans;
}

int main() {
    scanf( "%lld%lld", &n, &s );
    for( long long i = 1; i <= n; ++i ) scanf( "%lld", &f[ i ] );
    long long Ans = 0;
    for( long long i = 0; i < 1 << n; ++i ) {
        long long t, Cnt = 0, Pos = s;
        for( t = i; t; t >>= 1 ) if( t & 1 ) ++Cnt;
        for( long long j = 1, t = i; t; t >>= 1, ++j ) if( t & 1 ) Pos -= f[ j ] + 1;
        if( Pos < 0 ) continue;
        Ans += ( Cnt & 1 ) ? -C( Pos + n - 1, n - 1 ) : C( Pos + n - 1, n - 1 );
        Ans = ( Ans + Mod ) % Mod;
    }
    printf( "%lld\n", Ans );
    return 0;
}