51nod 1597 有限背包计数问题 (背包 分块)

#include<cstdio>
#include<cmath>
#include<cstring>
#define pt(x) printf("%d\n", x);
using namespace std;
const int MAXN = 1e5 + 10, mod = 23333333;
int N, M, f[81][MAXN], g[81][MAXN];
int add(int x, int y) {
    return (x + y >= mod) ? (x + y - mod): x + y;
}
int main() {
    scanf("%d", &N);
    M = sqrt(N);

    /*f[0][0] = 1; int o = 1; 
    for(int i = 1; i <= M; i++) {
        for(int k = 0; k < i; k++) {//res
            int s = 0;
            for(int t = 0; i * t + k <= N; t++) {//num
                s = add(s, f[i - 1][k + t * i]);
                f[i][k + t * i] = s;
                if(t >= i) s = (s - f[i - 1][(t - i) * i + k] + mod) % mod;//over take
            }
        }
    }   
    int ans = f[M][N];
    
    pt(ans)
    
    g[0][0] = 1; int p = 0;
    
    for(int i = 1; i <= M; i++) {// used i goods
        for(int j = 0; j <= N; j++) {// length is j     
            if(j >= M + 1) g[i][j] = g[i - 1][j - (M + 1)];
            if(j >= i)     g[i][j] = add(g[i][j], g[i][j - i]);
        }
        for(int j = 0; j <= N; j++) (ans += 1ll * f[M][j] * g[i][N - j] % mod) %= mod;
    }
    printf("%d", ans);*/
    
    f[0][0] = 1; int o = 1; 
    for(int i = 1; i <= M; i++, o ^= 1) {
        memset(f[o], 0, sizeof(f[o]));
        for(int k = 0; k < i; k++) {//res
            int s = 0;
            for(int t = 0; i * t + k <= N; t++) {//num
                s = add(s, f[o ^ 1][k + t * i]);
                f[o][k + t * i] = s;
                if(t >= i) s = (s - f[o ^ 1][(t - i) * i + k] + mod) % mod;//over take
            }
        }
    }   
    int ans = f[o ^ 1][N], tmp = o ^ 1; 
    
    pt(ans)
    g[0][0] = 1; o = 1;
    for(int i = 1; i <= M; i++, o ^= 1) {// used i goods
        memset(g[o], 0, sizeof(g[o]));
        for(int j = 0; j <= N; j++) {// length is j     
            if(j >= M + 1) g[o][j] = g[o ^ 1][j - (M + 1)];
            if(j >= i)     g[o][j] = add(g[o][j], g[o][j - i]);
        }
        for(int j = 0; j <= N; j++) (ans += 1ll * f[tmp][j] * g[o][N - j] % mod) %= mod;
    }
    printf("%d", ans);
    return 0;
}