前所未见的思路,对于查询的Y的规模不同,用不同的查找方式,如果Y大的话就用线段树进行分段查找,小的话就直接线性查找了。时间给的10s还是很充裕的。
这就说明了,现场赛的时候要大胆搞,说不定就能过
#include <cstdio> #include <cstring> #include <algorithm> #include <map> #include <set> #include <bitset> #include <queue> #include <stack> #include <string> #include <iostream> #include <cmath> #include <climits> using namespace std; const int maxn = 5e5 + 10; const int inf = 1e9; const int MAX = 5e5; #define lson rt << 1, l, mid #define rson rt << 1 | 1, mid + 1, r int minv[maxn << 2], num[maxn], numcnt, n, itime[maxn]; int maxval; void build(int rt, int l, int r) { int mid = (l + r) >> 1; minv[rt] = inf; if(l == r) return; build(lson); build(rson); } void update(int rt, int l, int r, int pos, int val) { if(l == r) minv[rt] = val; else { int mid = (l + r) >> 1; if(pos <= mid) update(lson, pos, val); else update(rson, pos, val); minv[rt] = min(minv[rt << 1], minv[rt << 1 | 1]); } } int query(int rt, int l, int r, int ql, int qr) { if(ql <= l && qr >= r) return minv[rt]; else { int mid = (l + r) >> 1, ret = inf; if(ql <= mid) ret = min(ret, query(lson, ql, qr)); if(qr > mid) ret = min(ret, query(rson, ql, qr)); return ret; } } int ask_min(int mod) { int ans = inf, anstime = -1; for(int i = numcnt - 1; i >= 0; i--) { if(num[i] % mod < ans) { ans = num[i] % mod; anstime = itime[num[i]]; } if(ans == 0) break; } return anstime; } int ask_max(int mod) { int l = 0, r = mod - 1, ans = inf, anstime = -1; while(l <= MAX) { if(r > MAX) r = MAX; int nowval = query(1, 0, MAX, l, r); if(nowval % mod == ans && nowval != inf) { anstime = max(anstime, itime[nowval]); } if(nowval % mod < ans && nowval != inf) { ans = nowval % mod; anstime = itime[nowval]; } l += mod; r += mod; } return anstime; } int main() { int kase = 1; while(scanf("%d", &n) && n) { if(kase > 1) puts(""); numcnt = 0; maxval = 0; printf("Case %d: ", kase++); if(n == 0) break; build(1, 0, MAX); char cmd[10]; int val; for(int i = 1; i <= n; i++) { scanf("%s%d", cmd, &val); if(cmd[0] == 'B') { update(1, 0, MAX, val, val); num[numcnt++] = val; itime[val] = numcnt; maxval = max(maxval, val); } else { int ret; if(val <= 5000) ret = ask_min(val); else ret = ask_max(val); printf("%d ", ret); } } } return 0; }