线段树的运用 线段树真np 真好用，真强
题目链接
https://ac.nowcoder.com/acm/contest/3005/C
直接扫一遍维护区间 会出现 遇见 0 处理麻烦的问题
直接用线段树 这一问题将会不复存在 将普通的区间加 变成 区间 * 就可！
#include #include #include #include #include #include #include #include #define IO ios::sync_with_stdio(false), cin.tie(0), cout.tie(0); using namespace std; typedef long long LL; int dis[4][2] = {0, 1, 1, 0, 0, -1, -1, 0}; const int maxn = 200000 + 10; const LL mod = 998244353; const int inf = 0x3f3f3f3f; LL a[maxn]; LL c[maxn << 2]; void creat(int node, int L, int R) { if (L == R) { c[node] = a[L]; return; } int mid = (L + R) >> 1; creat(node << 1, L, mid); creat(node << 1 | 1, mid + 1, R); c[node] = c[node << 1 | 1] * c[node << 1] % mod; } LL query(int node, int L, int R, int start, int end) // L, R 为查询区间 { if (start > R || end < L) { return 1; } if (start >= L && end <= R) { return c[node]; } else { int mid = (start + end) >> 1; LL lans = 1; LL rans = 1; if (mid >= L) lans = query(node << 1, L, R, start, mid) % mod; if (mid + 1 <= R) rans = query(node << 1 | 1, L, R, mid + 1, end) % mod; return (lans * rans) % mod; } } int main() { #ifdef ONLINE_JUDGE #else freopen("in.txt", "r", stdin); // freopen("out.txt", "w", stdout); #endif IO; int n, k; LL ans = 0; cin >> n >> k; for (int i = 1; i <= n; i++) cin >> a[i]; creat(1, 1, n); for (int i = 1; i + k - 1 <= n; i++) { int x = i; int y = i + k - 1; ans = max(ans, query(1, x, y, 1, n) % mod) % mod; } cout << ans; return 0; }
