• POJ 3145 Harmony Forever 线段树


    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;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/rolight/p/4073339.html
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