思路
考虑每一位对于答案的影响，先枚举多少个 1，然后考虑每一位可以产生多少个贡献，那么我们就可以得到答案为：\(\sum_{i=0}^{m}\sum_{i=0}^{n}C_{n}^{i}2^i\) 把 \(2^{i}\) 提取出来就是 \(2^{n+1}-1\) 然后你就会发现剩下的是一个二次项系数的一部分，然后你就可以用莫队维护这个，不过你还可以用分块快速求解，这里给出分块预处理后在线查值，思路太短了，放个图凑数：
#include <bits/stdc++.h>
#define int long longusing namespace std;const int MaxN = 2e5 + 10, mod = 998244353, B = 447;struct Block {vector<int> p2;vector<vector<int>> d;int f[MaxN], v[MaxN], n;int qpow(int a, int b, int res = 1) {for (int i = 1; i <= b; i <<= 1) {(b & i) && (res = res * a % mod);a = a * a % mod;}return res;}void init() {f[0] = 1;for (int i = 1; i < MaxN; ++i) {f[i] = f[i - 1] * i % mod;}v[MaxN - 1] = qpow(f[MaxN - 1], mod - 2);for (int i = MaxN - 2; i >= 0; --i) {v[i] = v[i + 1] * (i + 1) % mod;}}int g(int x, int k) {int m = k * B;return x <= m ? p2[x] : d[k][x - m] * v[m] % mod;}Block(int _n) : n(_n) {p2.resize(n + 1), p2[0] = 1, init();for (int i = 1; i <= n; ++i) {p2[i] = (p2[i - 1] * 2) % mod;}int K = (n + B - 1) / B + 2;d.resize(K);for (int k = 0; k < K; ++k) {int m = k * B;if (m > n) continue;d[k].resize(n - m + 1);d[k][0] = p2[m] * f[m] % mod;for (int i = 0; i < n - m; ++i) {d[k][i + 1] = ((d[k][i] * 2 % mod - f[i + m] * v[i] % mod) % mod + mod) % mod;}}}int query(int x, int m, int a = 0, int res = 0) {m = min(m, x + 1);if (m <= 0) return 0;if (m + m > x + 1) return (p2[x] - query(x, x + 1 - m) + mod) % mod;m--, a = m / B;if (m <= a * B + B / 2) {for (int i = a * B + 1; i <= m; ++i) {res = (res + v[i] * v[x - i] % mod) % mod;}return (g(x, a) + res * f[x] % mod) % mod;}for (int i = m + 1; i <= min((a + 1) * B, x); ++i) {res = (res + v[i] * v[x - i] % mod) % mod;}return (g(x, a + 1) - res * f[x] % mod + mod) % mod;}
};int t, n, m;signed main() {Block B(2e5 + 1);for (cin >> t; t; t--) {cin >> n >> m;cout << ((B.qpow(2, n) - 1) % mod + mod) % mod * B.query(n - 1, m) % mod << '\n';}return 0;
}