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注意1的情况,顺便粘一发代码#include<cstdio> #include<string> #include<cstring> #include<cstdlib> #include<iostream> #include<algorithm> using namespace std; const int N = 20005; int n, ans = 1, prime = 1; int pre[N], pri[N], tot = 0; signed main(void) { memset(pre, 0, sizeof(pre)); for (int i = 2; i < N; i++) { if (pre[i] == 0)pri[++tot] = i; for (int j = 1; j <= tot&&i*pri[j] < N; j++) { pre[i*pri[j]] = pri[j]; if (i%pri[j] == 0)break; } } scanf_s("%d", &n); for (int i = 1, x; i <= n; i++) { scanf_s("%d", &x); for(int j=lower_bound(pri+1,pri+1+tot,x)-pri;j;j--) if (pri[j] <= x&&x%pri[j] == 0) { if (prime < pri[j])prime = pri[j], ans = x; break; } } printf("%d\n", ans); } Followed by: Post your reply here: |
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