Codeforces Round #253 (Div. 1) (A, B, C)
2014-09-04 12:17
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Codeforces Round #253 (Div. 1)
题目链接A:给定一些牌,然后如今要提示一些牌的信息,要求提示最少,使得全部牌能够被分辨出来。
思路:一共2^10种情况,直接暴力枚举,然后对于每种情况去推断,推断的时候仅仅要两两张牌枚举出来推断就可以。不得不说CF机子真心强大,2秒限制还是能跑10^8
B:给定n个发生概率,求选出当中一些事件,使得正好有一件发生的概率最大。
思路:贪心,从大到小排序概率,然后一个个概率进来推断有没有更大,有就增加该事件,没有就跳过
C:给定n个数字,要求一个个删掉数字,假设数字左边或右边没有数字就不得分,否则得分为min(左边数字,右边数字)。
思路:贪心,考虑两种情况,一种下凹的形状,这样的情况下肯定是由中间的优先拿掉在到两边,对于这样的情况能够用一个栈去维护,每次要入栈前,先把之间拿掉并记录答案,在入栈。
这样处理完了,下凹的形状都没了,图形为先递增后递减形状,这时候答案是固定的了,就是最大的两个值肯定是取不到了,剩下的都能取到。这两部分和加起来就是答案
代码:
A题:
#include <stdio.h> #include <string.h> #include <algorithm> using namespace std; const char c[15] = {'1', '2', '3', '4', '5', 'B', 'Y', 'W', 'G', 'R'}; int n, vis[105][2]; char card[105][5]; int bitcount(int x) { if (x == 0) return 0; return bitcount(x>>1) + (x&1); } bool judge(int s) { memset(vis, 0, sizeof(vis)); for (int i = 0; i < 10; i++) { if (s&(1<<i)) { for (int j = 0; j < n; j++) { for (int k = 0; k < 2; k++) { if (card[j][k] == c[i]) vis[j][k] = 1; } } } } for (int i = 0; i < n; i++) { if (vis[i][0] && vis[i][1]) continue; for (int j = i + 1; j < n; j++) { if (strcmp(card[i], card[j]) == 0) continue; if (vis[j][0] && vis[j][1]) continue; if (!vis[i][0] && !vis[i][1] && !vis[j][0] && !vis[j][1]) return false; if (!vis[i][0] && !vis[j][0] && card[i][1] == card[j][1]) return false; if (!vis[i][1] && !vis[j][1] && card[i][0] == card[j][0]) return false; } } return true; } int main() { scanf("%d", &n); for (int i = 0; i < n; i++) scanf("%s", card[i]); int ans = 10; for (int i = 0; i < (1<<10); i++) { if (judge(i)) { ans = min(ans, bitcount(i)); } } printf("%d\n", ans); return 0; }
B题:
#include <stdio.h> #include <string.h> #include <algorithm> using namespace std; const int N = 105; int n, save , sn; double p , sum; double cal(int now) { double ss = sum * (1 - p[now]); double ans = sum * p[now]; for (int i = 0; i < sn; i++) { ans += ss / (1 - p[save[i]]) * p[save[i]]; } return ans; } int main() { scanf("%d", &n); for (int i = 0; i < n; i++) { scanf("%lf", &p[i]); } sort(p, p + n); sn = 1; save[0] = n - 1; sum = (1 - p[n - 1]); double Max = p[n - 1]; for (int i = n - 2; i >= 0; i--) { double tmp = cal(i); if (tmp > Max) { sum *= (1 - p[i]); Max = tmp; save[sn++] = i; } } printf("%.12lf\n", Max); return 0; }
C题:
#include <stdio.h> #include <string.h> #include <algorithm> using namespace std; const int N = 500005; int n, top; long long stack , x, ans; int main() { scanf("%d", &n); for (int i = 0; i < n; i++) { scanf("%d", &x); while (top > 0 && stack[top - 1] >= stack[top] && x >= stack[top]) { ans += min(stack[top - 1], x); top--; } stack[++top] = x; } sort(stack + 1, stack + 1 + top); for (int i = 1; i <= top - 2; i++) { ans += stack[i]; } printf("%lld\n", ans); return 0; }
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