比赛征程
| # | A | B | C | D | E |
|---|---|---|---|---|---|
| contest19 | Y | A | Y | A | A |
A - Adding Digits
Problem Description
给出a和b，要求给a右端加一位数字n次且次次是b的倍数。
Time Limit: 2s
Solution
第一次扩展如果成功，那么之后的扩展全部是0就能除得尽。
B - Ancient Prophesy
注意要认真审题.
C - Balls and Boxes
Problem Description
编号1-n个桶，每个桶里有ai个球，现在从下标为i的桶里，拿出所有的球，一直往右边放，放到n了，就再从0开始放，告诉你最后一个球放下的下标，要你输出所有桶的最初状态。
Time Limit: 2s
Solution
强行模拟, 用前缀后缀优化时间
D - Black and White Tree
Problem Description
有n个点的一个树，同一条边的两个点涂成不同的颜色(black & white)，每条边有一个权值，题目给出n ， 和每个点的颜色 和 与每个点相连的边的权值和。
Time Limit: 2s
Solution
按s从小到大排序贪心地建边，并消耗完s。由于黑点和白点的权值和相同,就可以不断的贪心消除.
Code
#include <cstdio> #include <iostream> #include <algorithm> using namespace std; const int N = 100005; struct node{ int s, c, id; }tree[N]; int cw, cb; inline bool cmp(node a, node b) { return a.c < b.c; } int main() { int n, c, s, pw = 1, pb = 1, t; scanf("%d", &n); for (int i = 1; i <= n; i ++) scanf("%d%d", &tree[i].c, &tree[i].s), tree[i].id = i; sort(tree + 1, tree + n + 1, cmp); for (pb = 1; pb <= n && !tree[pb].c; pb ++); for (int i = 1; i < n; i ++) { t = min(tree[pb].s, tree[pw].s); printf("%d %d %d\n", tree[pb].id, tree[pw].id, t); tree[pb].s -= t; tree[pw].s -= t; if (pb < n && !tree[pb].s) pb ++; else pw ++; } return 0; }
E - Dividing Kingdom
Problem Description
给出一些点，要求给出4条线，两条平行x轴，两条平等y轴，不经过任何 点，把平面分为9块，每块包含的点数，正数可以满足每个人的需要
Time Limit: 2s
Solution
用主席树判断每个区间是否存在符合要求的方案, 然后看每个区间方案是否能有交集
Code
#include <cstdio> #include <iostream> #include <algorithm> using namespace std; typedef double db; typedef pair<int, int> P; const int N = 100005; const int M = 15000000; P p[N]; int tab[N]; int lc[M], rc[M], key[M], cnt = 0; int root[N], rtab[N], tt[N]; int pos; bool ft; inline void add(int x, int y, int l, int r, int pos) { key[x] = key[y] + 1; if (l == r) return; int mid = (l + r) >> 1; if (pos <= mid) { if (!lc[x]) lc[x] = ++ cnt; rc[x] = rc[y]; add(lc[x], lc[y], l, mid, pos); } else { if (!rc[x]) rc[x] = ++ cnt; lc[x] = lc[y]; add(rc[x], rc[y], mid + 1, r, pos); } } inline void dfs(int x, int y, int l, int r, int len) { if (l == r) { pos = l; if (len == key[x] - key[y]) ft = 1; return; } int mid = (l + r) >> 1; if (len <= key[lc[x]] - key[lc[y]]) dfs(lc[x], lc[y], l, mid, len); else dfs(rc[x], rc[y], mid + 1, r, len - (key[lc[x]] - key[lc[y]])); } int main() { int n, x, y; scanf("%d", &n); for (int i = 1; i <= n; i ++) { scanf("%d%d", &x, &y); p[i] = make_pair(x, y); tab[i] = y; } sort(p + 1, p + n + 1); sort(tab + 1, tab + n + 1); int tot = unique(tab + 1, tab + n + 1) - tab - 1, t = 0, cur, lst; for (int i = 1, j = 1; i <= n; i = j + 1) { j = i; while (j < n && p[j + 1].first == p[i].first) j ++; lst = root[t ++]; tt[t] = p[i].first; for (int k = i; k <= j; k ++) { p[k].second = lower_bound(tab + 1, tab + tot + 1, p[k].second) - tab; cur = ++ cnt; add(cur, lst, 1, tot, p[k].second); lst = cur; } root[t] = cur; rtab[t] = key[cur]; } int a[11], tmp[11]; for (int i = 1; i <= 9; i ++) scanf("%d", &a[i]); sort(a + 1, a + 10); int p, q, rd, sd, ru, su; bool flag; while (true) { flag = 1; tmp[0] = 0; for (int i = 1; i <= 9; i ++) tmp[i] = tmp[i - 1] + a[i]; p = lower_bound(rtab + 1, rtab + t + 1, tmp[3]) - rtab; flag = (flag && rtab[p] == tmp[3]); q = lower_bound(rtab + 1, rtab + t + 1, tmp[6]) - rtab; flag = (flag && rtab[q] == tmp[6]); ft = 0; dfs(root[p], 0, 1, tot, a[1]); rd = pos; flag = (flag && ft); dfs(root[p], 0, 1, tot, a[1] + 1); ru = pos; ft = 0; dfs(root[p], 0, 1, tot, a[1] + a[2]); sd = pos; flag = (flag && ft); dfs(root[p], 0, 1, tot, a[1] + a[2] + 1); su = pos; ft = 0; dfs(root[q], root[p], 1, tot, a[4]); rd = max(rd, pos); flag = (flag && ft); dfs(root[q], root[p], 1, tot, a[4] + 1); ru = min(ru, pos); ft = 0; dfs(root[q], root[p], 1, tot, a[4] + a[5]); sd = max(sd, pos); flag = (flag && ft); dfs(root[q], root[p], 1, tot, a[4] + a[5] + 1); su = min(su, pos); ft = 0; dfs(root[t], root[q], 1, tot, a[7]); rd = max(rd, pos); flag = (flag && ft); dfs(root[t], root[q], 1, tot, a[7] + 1); ru = min(ru, pos); ft = 0; dfs(root[t], root[q], 1, tot, a[7] + a[8]); sd = max(sd, pos); flag = (flag && ft); dfs(root[t], root[q], 1, tot, a[7] + a[8] + 1); su = min(su, pos); if (flag && sd < su && rd < ru) { printf("%.10lf %.10lf\n%.10lf %.10lf\n", (db)tt[p] + 0.5, (db)tt[q] + 0.5, (db)tab[rd] + 0.5, (db)tab[sd] + 0.5); return 0; } if (!next_permutation(a + 1, a + 10)) break; } puts("-1"); return 0; }