送分题 [组合计数]

$送分题$

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$color{red}{正解部分}$

考虑每一位对答案的贡献, 在长度为 $N$ 的序列中, 现在计算第 $bit$ 位的答案,
设 $bit$ 位 为 $1$ 的数量为 $n_1$, 为 $0$ 的数量为 $n_0$, $x+y = N$,

对答案的贡献为 $left(sumlimits_{i=1}^{n_1} [i\%2 =1]egin{pmatrix} N \ i end{pmatrix}x^i ight) imes left( sumlimits_{i=0}^{n_0}egin{pmatrix} N \ i end{pmatrix}x^i ight)$ .

根据 二项式定理 化第二项为: $(x+1)^{n_0}$ .

将第一项化为: $frac{sumlimits_{i=0}^{n_1} left((-1)^{i+1}+1 ight) egin{pmatrix} N \ i end{pmatrix}x^i}{2}$

将括号拆开得到: $frac{-sumlimits_{i=0}^{n_1}egin{pmatrix} N \ iend{pmatrix}(-x)^i+sumlimits_{i=0}^{n_1} egin{pmatrix} N \ iend{pmatrix}x^i}{2}$

再使用 二项式定理 继续化简得到: $frac{-(-x+1)^{n_1} + (x+1)^{n_1}}{2}$

$herefore ans = sumlimits_{bit=0}^{30} frac{-(-x+1)^{n_1} + (x+1)^{n_1}}{2} imes (x+1)^{n_0} imes 2^{bit}$ .

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$color{red}{实现部分}$

#include<bits/stdc++.h>
#define reg register

int read(){
        char c;
        int s = 0, flag = 1;
        while((c=getchar()) && !isdigit(c))
                if(c == '-'){ flag = -1, c = getchar(); break ; }
        while(isdigit(c)) s = s*10 + c-'0', c = getchar();
        return s * flag;
}

const int maxn = 1e5 + 10;
const int mod = 998244353;

int N;
int M;
int A[maxn];
int pw[maxn];
int s[32][maxn];

int Ksm(int a, int b){ int s=1; while(b){if(b&1)s=1ll*s*a%mod;a=1ll*a*a%mod;b>>=1;} return s; }

void Work(){
        int l = read(), r = read(), x = read(), Ans = 0;
        for(reg int bit = 30; bit >= 0; bit --){
                int n1 = s[bit][r] - s[bit][l-1], n0 = r-l+1-n1;
                int res = (Ksm(x+1, n1) - Ksm((1-x+mod)%mod, n1) + mod) % mod;
                res = 1ll*res*Ksm(2, mod-2) % mod;
                res = 1ll*res * Ksm(x+1, n0) % mod;
                res = 1ll*res*pw[bit] % mod;
                Ans = (Ans + res) % mod;
        }
        printf("%d
", Ans);
}

int main(){
        N = read(), M = read();
        for(reg int i = 1; i <= N; i ++) A[i] = read();
        pw[0] = 1; for(reg int i = 1; i <= 30; i ++) pw[i] = 2ll*pw[i-1] % mod;
        for(reg int i = 0; i <= 30; i ++)
                for(reg int j = 1; j <= N; j ++){
                        if(A[j] & (1 << i)) s[i][j] ++;
                        s[i][j] += s[i][j-1];
                }
        for(reg int i = 1; i <= M; i ++) Work();
        return 0;
}