A. Toda 2
题意:给你n个人,每个人的分数是a[i],每次可以从两个人到五个人的使得分数减一,使得最终的分数相等;
思路:假设答案为m;每个人的分数与答案m的差值为d[i],sum为d[i]的总和,max为d[i]的最大值;仅当sum-max>=max的时候才满足;
满足之后,总和为奇数,先取三个,再取两个(都是最大与次大值);偶数每次取两个即可;
B. Minimum and Maximum
题意:人机交互题;给你一个数组 ,找出其中的最小值与最大值,需要在询问f(n)的次数内得到最大值与最小值;
每次询问,? i j表示i和j的下标,告诉你哪个大,相等或者小,得到答案;
思路:类似与归并排序的思想,每次取两个,得到最大最小值;
G. Car Repair Shop
题意:给你n个区间,起点和长度;当当前区间与前面的区间有交集的时候,找到另外一个相同长度的区间放进去;
思路:模拟,先有一个大区间,每次插入一个区间,把这个区间修改成空白点的,把那个大区间拆成两个小区间,如果原来的区间内有空白,从1开始查找是否有长度
大于当前长度的区间,放入;
set模拟;
H. Delete Them
队友写的;
#include<bits/stdc++.h> using namespace std; #pragma comment(linker, "/STACK:102400000,102400000") #define ls i<<1 #define rs ls | 1 #define mid ((ll+rr)>>1) #define pii pair<int,int> #define MP make_pair typedef long long LL; const long long INF = 1e18+1LL; const double Pi = acos(-1.0); const int N = 1e3+10, M = 2e5+20, mod = 1e9+7, inf = 2e9; int a[N],n,m,pos[N],H[N]; char s[N][N]; set<int > S; int main() { scanf("%d%d",&n,&m); for(int i = 1; i <= n; ++i) scanf("%s",s[i]); for(int i = 1; i <= m; ++i) scanf("%d",&pos[i]),H[pos[i]]=1; sort(pos+1,pos+m+1); for(int i = 1; i <= m; ++i) S.insert(strlen(s[pos[i]])); if(S.size() > 1) { puts("No"); return 0; } for(int i = 2; i <= m; ++i) { for(int j=0;j<strlen(s[pos[i]]); ++j) { if(s[pos[1]][j]!=s[pos[i]][j]) { s[pos[1]][j] = '?'; } } } for(int i = 1;i<=n; ++i) if(!H[i]) { int len = strlen(s[i]); int sum = 0; if(len!=strlen(s[pos[1]])) continue; for(int j = 0; j < len; ++j) { if(s[pos[1]][j] == '?') sum++; else if(s[pos[1]][j] == s[i][j]) sum++; } if(sum == len) { puts("No"); return 0; } } puts("Yes"); printf("%s\n",s[pos[1]]); }
J. Bottles
题意:给你n个瓶子,第一行a[i]表示第i个瓶子的剩余水的容量,第二行b[i]表示第i个瓶子的体积;
你需要使用最少的瓶子数量,装完剩下全部的水,每次改变n单位的水的花费n;
思路:k很好求,贪心求法,开始以为n就100,写个暴搜+剪枝,TLE TEST 42;
后面发现就一个背包,只是这个背包需要求的是恰好装满这个背包大小的最大价值;
剩余水的质量为sum1,瓶子的体积和sum2;你需要找到sum1-sum2之间的最大值;
dp[k][sum1]-dp[k][sum2];
复杂度(k*sum2*n);
优化下可以(k*sum1*n);
Nick has n bottles of soda left after his birthday. Each bottle is described by two values: remaining amount of soda ai and bottle volumebi (ai ≤ bi).
Nick has decided to pour all remaining soda into minimal number of bottles, moreover he has to do it as soon as possible. Nick spends xseconds to pour x units of soda from one bottle to another.
Nick asks you to help him to determine k — the minimal number of bottles to store all remaining soda and t — the minimal time to pour soda into k bottles. A bottle can't store more soda than its volume. All remaining soda should be saved.
The first line contains positive integer n (1 ≤ n ≤ 100) — the number of bottles.
The second line contains n positive integers a1, a2, ..., an (1 ≤ ai ≤ 100), where ai is the amount of soda remaining in the i-th bottle.
The third line contains n positive integers b1, b2, ..., bn (1 ≤ bi ≤ 100), where bi is the volume of the i-th bottle.
It is guaranteed that ai ≤ bi for any i.
The only line should contain two integers k and t, where k is the minimal number of bottles that can store all the soda and t is the minimal time to pour the soda into k bottles.
4
3 3 4 3
4 7 6 5
2 6
2
1 1
100 100
1 1
5
10 30 5 6 24
10 41 7 8 24
3 11
In the first example Nick can pour soda from the first bottle to the second bottle. It will take 3 seconds. After it the second bottle will contain 3 + 3 = 6 units of soda. Then he can pour soda from the fourth bottle to the second bottle and to the third bottle: one unit to the second and two units to the third. It will take 1 + 2 = 3 seconds. So, all the soda will be in two bottles and he will spend 3 + 3 = 6seconds to do it.