\(\mathcal{Description}\)

  Link.

  有 \(n\) 种物品，第 \(i\) 中有 \(a_i\) 个，单价为 \(b_i\)。共 \(q\) 次询问，每次查询用不超过 \(c\) 的钱购买种类在 \([l,r]\) 之中的物品，有多少种方案。强制在线；答案对 \(998244353\) 取模。

  \(n\le10^4\)，\(q\le5\times10^4\)，\(c\le10^3\)。

\(\mathcal{Solution}\)

  快速回答区间询问，最基础但容易被忽略的处理方式——前缀和差。

  考虑第 \(i\) 中物品的 OGF，显然有

\[G_i(x)=\frac{1-x^{(a_i+1)b_i}}{1-x^{b_i}}. \]

欲求答案 \(\sum_{k\le c}[x^k]\prod_{i=l}^rG_i(x)\)，转化为前缀积乘上前缀积的逆，预处理出

\[S_i(x)=\prod_{j=1}^iG_j(x),\\ S^{-1}_i(x)=\prod_{j=1}^iG_j^{-1}(x). \]

顺带发现 \(G_j(x)\) 和 \(G_j^{-1}\) 长相完全一样，所以这俩也就是换换加减号的事儿。精巧递推一发可以做到 \(\mathcal O(nc)\) 预处理，查询复杂度即求前缀系数和，预先将 \(S_i(x)\) 或 \(S_i^{-1}(x)\) 的系数做前缀和后即为求卷积的某项系数，暴力模拟，则有单次查询复杂度 \(\mathcal O(c)\)。

\(\mathcal{Code}\)

/* Clearink */

#include <cstdio>
#include <cstring>

#define rep( i, l, r ) for ( int i = l, rep##i = r; i <= rep##i; ++i )
#define per( i, r, l ) for ( int i = r, per##i = l; i >= per##i; --i )

inline int rint() {
    int x = 0, s = getchar();
    for ( ; s < '0' || '9' < s; s = getchar() );
    for ( ; '0' <= s && s <= '9'; s = getchar() ) x = x * 10 + ( s ^ '0' );
    return x;
}

inline void wint( const int x ) {
    if ( 9 < x ) wint( x / 10 );
    putchar( x % 10 ^ '0' );
}

const int MAXN = 1e4, MAXC = 1e3, MOD = 998244353;
int n, q, a[MAXN + 5], b[MAXN + 5];
int f[MAXN + 5][MAXC + 5], g[MAXN + 5][MAXC + 5];

inline void subeq( int& a, const int b ) { ( a -= b ) < 0 && ( a += MOD ); }
inline int sub( int a, const int b ) { return ( a -=  b ) < 0 ? a + MOD : a; }
inline void addeq( int& a, const int b ) { ( a += b ) >= MOD && ( a -= MOD ); }
inline int add( int a, const int b ) { return ( a += b ) < MOD ? a : a - MOD; }
inline int mul( const long long a, const int b ) { return int( a * b % MOD ); }

inline void init() {
    f[0][0] = g[0][0] = 1;
    rep ( i, 1, n ) {
        memcpy( f[i], f[i - 1], sizeof f[i] );
        memcpy( g[i], g[i - 1], sizeof g[i] );
        int t;
        rep ( j, t = b[i], MAXC ) addeq( f[i][j], f[i][j - t] );
        per ( j, MAXC, t = ( a[i] + 1 ) * b[i] ) subeq( f[i][j], f[i][j - t] );
        rep ( j, t = ( a[i] + 1 ) * b[i], MAXC ) addeq( g[i][j], g[i][j - t] );
        per ( j, MAXC, t = b[i] ) subeq( g[i][j], g[i][j - t] );
    }
    rep ( i, 0, n ) rep ( j, 1, MAXC ) addeq( g[i][j], g[i][j - 1] );
}

int main() {
    freopen( "shop.in", "r", stdin );
    freopen( "shop.out", "w", stdout );

    n = rint(), q = rint();
    rep ( i, 1, n ) a[i] = rint(), b[i] = rint();

    init();

    for ( int ans = 0, l, r, c; q--; ) {
        l = ( rint() + ans ) % n + 1, r = ( rint() + ans ) % n + 1, c = rint();
        if ( l > r ) l ^= r ^= l ^= r;
        ans = 0;
        rep ( i, 0, c ) addeq( ans, mul( f[r][i], g[l - 1][c - i] ) );
        wint( ans ), putchar( '\n' );
    }
    return 0;
}

标签：int,rint,rep,Solution,多校,联训,ans,mathcal,MOD
来源： https://www.cnblogs.com/rainybunny/p/14915420.html