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Codeforces Round #393 (Div. 2)
Petr wants to make a calendar for current month. For this purpose he draws a table in which columns correspond to weeks (a week is seven consequent days from Monday to Sunday), rows correspond to weekdays, and cells contain dates. For example, a calendar for January 2017 should look like on the picture:
Petr wants to know how many columns his table should have given the month and the weekday of the first date of that month? Assume that the year is non-leap.
The only line contain two integers m and d (1 ≤ m ≤ 12, 1 ≤ d ≤ 7) — the number of month (January is the first month, December is the twelfth) and the weekday of the first date of this month (1 is Monday, 7 is Sunday).
Print single integer: the number of columns the table should have.
1 7
6
1 1
5
11 6
5
The first example corresponds to the January 2017 shown on the picture in the statements.
In the second example 1-st January is Monday, so the whole month fits into 5 columns.
In the third example 1-st November is Saturday and 5 columns is enough.
题意：
假设没有闰年，给出每个月第一天是星期几，问这个月的日历有几列...
代码：
#include<algorithm>#include<iostream>#include<cstring>#include<cstdio>//by NeighThornusing namespace std;int n,m,ans=0;inline int judge(int x){ if(x<=7){ if(x&1) return 31; else if(x==2) return 28; else return 30; } else{ if(x&1) return 30; return 31; }}signed main(void){ scanf("%d%d",&n,&m); n=judge(n); n-=7-m+1; ans=1+(n+6)/7; printf("%d\n",ans); return 0;}//Cap ou pas cap. Cap.n hobbits are planning to spend the night at Frodo‘s house. Frodo has n beds standing in a row and m pillows (n ≤ m). Each hobbit needs a bed and at least one pillow to sleep, however, everyone wants as many pillows as possible. Of course, it‘s not always possible to share pillows equally, but any hobbit gets hurt if he has at least two pillows less than some of his neighbors have.
Frodo will sleep on the k-th bed in the row. What is the maximum number of pillows he can have so that every hobbit has at least one pillow, every pillow is given to some hobbit and no one is hurt?
The only line contain three integers n, m and k (1 ≤ n ≤ m ≤ 109, 1 ≤ k ≤ n) — the number of hobbits, the number of pillows and the number of Frodo‘s bed.
Print single integer — the maximum number of pillows Frodo can have so that no one is hurt.
4 6 2
2
3 10 3
4
3 6 1
3
In the first example Frodo can have at most two pillows. In this case, he can give two pillows to the hobbit on the first bed, and one pillow to each of the hobbits on the third and the fourth beds.
In the second example Frodo can take at most four pillows, giving three pillows to each of the others.
In the third example Frodo can take three pillows, giving two pillows to the hobbit in the middle and one pillow to the hobbit on the third bed.
题意：
有n张床排成一行，Frodo在第k张床，共有m个枕头，每个人至少有1个枕头，并且每个人相邻的两个人只能比他多至多一个枕头，问Frodo最多可以得到多少个枕头...
分析：
二分答案...
最优解一定是阶梯型的...
假设Frodo的一边有y张床，当前二分的ans=x，如果y大于x-1，那么这一边加上Frodo最少共有$\frac{(x+1)x}{2}+y-(x-1)$个枕头，否则就是$\frac{(x+x-y)(y+1)}{2}$...然后判断就好了...
代码：
#include<algorithm>#include<iostream>#include<cstring>#include<cstdio>//by NeighThornusing namespace std;int n,m,k;inline long long calc(int x,int y){ if(!y) return x; if(!x) return 0; if(y>x-1) return 1LL*(x+1)*x/2LL+y-(x-1); return 1LL*(x+x-y)*(y+1)/2LL;}signed main(void){ scanf("%d%d%d",&n,&m,&k); int l=0,r=1e9,ans=0; while(l<=r){ int mid=(l+r)>>1; if(calc(mid,k-1)+calc(mid,n-k)-mid<=m) ans=mid,l=mid+1; else r=mid-1; } printf("%d\n",ans); return 0;}//Cap ou pas cap. Cap.
C. Pavel and barbecue
Pavel cooks barbecue. There are n skewers, they lay on a brazier in a row, each on one of n positions. Pavel wants each skewer to be cooked some time in every of n positions in two directions: in the one it was directed originally and in the reversed direction.
Pavel has a plan: a permutation p and a sequence b1, b2, ..., bn, consisting of zeros and ones. Each second Pavel move skewer on position i to position pi, and if bi equals 1 then he reverses it. So he hope that every skewer will visit every position in both directions.
Unfortunately, not every pair of permutation p and sequence b suits Pavel. What is the minimum total number of elements in the given permutation p and the given sequence b he needs to change so that every skewer will visit each of 2n placements? Note that after changing the permutation should remain a permutation as well.
There is no problem for Pavel, if some skewer visits some of the placements several times before he ends to cook. In other words, a permutation p and a sequence b suit him if there is an integer k (k ≥ 2n), so that after k seconds each skewer visits each of the 2nplacements.
It can be shown that some suitable pair of permutation p and sequence b exists for any n.
The first line contain the integer n (1 ≤ n ≤ 2·105) — the number of skewers.
The second line contains a sequence of integers p1, p2, ..., pn (1 ≤ pi ≤ n) — the permutation, according to which Pavel wants to move the skewers.
The third line contains a sequence b1, b2, ..., bn consisting of zeros and ones, according to which Pavel wants to reverse the skewers.
Print single integer — the minimum total number of elements in the given permutation p and the given sequence b he needs to change so that every skewer will visit each of 2n placements.
4
4 3 2 1
0 1 1 1
2
3
2 3 1
0 0 0
1
In the first example Pavel can change the permutation to 4, 3, 1, 2.
In the second example Pavel can change any element of b to 1.
题意：
我真的没看懂题...现在都没看懂TAT...
A new innovative ticketing systems for public transport is introduced in Bytesburg. Now there is a single travel card for all transport. To make a trip a passenger scan his card and then he is charged according to the fare.
The fare is constructed in the following manner. There are three types of tickets:
- a ticket for one trip costs 20 byteland rubles,
- a ticket for 90 minutes costs 50 byteland rubles,
- a ticket for one day (1440 minutes) costs 120 byteland rubles.
Note that a ticket for x minutes activated at time t can be used for trips started in time range from t to t + x - 1, inclusive. Assume that all trips take exactly one minute.
To simplify the choice for the passenger, the system automatically chooses the optimal tickets. After each trip starts, the system analyses all the previous trips and the current trip and chooses a set of tickets for these trips with a minimum total cost. Let the minimum total cost of tickets to cover all trips from the first to the current is a, and the total sum charged before is b. Then the system charges the passenger the sum a - b.
You have to write a program that, for given trips made by a passenger, calculates the sum the passenger is charged after each trip.
The first line of input contains integer number n (1 ≤ n ≤ 105) — the number of trips made by passenger.
Each of the following n lines contains the time of trip ti (0 ≤ ti ≤ 109), measured in minutes from the time of starting the system. All ti are different, given in ascending order, i. e. ti + 1 > ti holds for all 1 ≤ i < n.
Output n integers. For each trip, print the sum the passenger is charged after it.
3
10
20
30
20
20
10
10
13
45
46
60
103
115
126
150
256
516
20
20
10
0
20
0
0
20
20
10
In the first example, the system works as follows: for the first and second trips it is cheaper to pay for two one-trip tickets, so each time 20 rubles is charged, after the third trip the system understands that it would be cheaper to buy a ticket for 90 minutes. This ticket costs 50 rubles, and the passenger had already paid 40 rubles, so it is necessary to charge 10 rubles only.
题意：
有三种车票可供选择：
No.0 坐一站，花费为20
No.1 坐90-1分钟，花费为50(因为在站台需要花费1分钟，所以要-1)
No.2 坐1440-1分钟，花费为120
给出n个站点，每一次给出到下一站点的时间，每一次买票都是最优选择，问每一次买票需要多交多少钱...
分析：
感觉这就是裸的DP然后输出每一次的结果减去上一次的结果...
代码：
#include<algorithm>#include<iostream>#include<cstring>#include<cstdio>//by NeighThornusing namespace std;const int maxn=100000+5;int n,t[maxn];long long f[maxn];signed main(void){ scanf("%d",&n); for(int i=1;i<=n;i++) scanf("%d",&t[i]); f[0]=0; for(int i=1;i<=n;i++){ f[i]=f[i-1]+20; long long a=lower_bound(t+1,t+n+1,t[i]-89 )-t-1,w1=20*(i-a+1); long long b=lower_bound(t+1,t+n+1,t[i]-1439)-t-1,w2=20*(i-b+1); if(w1>50 ) f[i]=min(f[i],f[a]+50 ); else f[i]=min(f[i],f[a]+w1); if(w2>120) f[i]=min(f[i],f[b]+120); else f[i]=min(f[i],f[b]+w2); printf("%I64d\n",f[i]-f[i-1]); } return 0;}//Cap ou pas cap. Cap.
By NeighThorn
Codeforces Round #393 (Div. 2)