A. Borya’s Diagnosis
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
It seems that Borya is seriously sick. He is going visit n doctors to find out the exact diagnosis. Each of the doctors needs the information about all previous visits, so Borya has to visit them in the prescribed order (i.e. Borya should first visit doctor 1, then doctor 2, then doctor 3 and so on). Borya will get the information about his health from the last doctor.Doctors have a strange working schedule. The doctor i goes to work on the si-th day and works every di day. So, he works on days si, si + di, si + 2di, ….
The doctor’s appointment takes quite a long time, so Borya can not see more than one doctor per day. What is the minimum time he needs to visit all doctors?
Input
First line contains an integer n — number of doctors (1 ≤ n ≤ 1000).Next n lines contain two numbers si and di (1 ≤ si, di ≤ 1000).
Output
Output a single integer — the minimum day at which Borya can visit the last doctor.Examples
input
3
2 2
1 2
2 2
output
4
input
2
10 1
6 5
output
11
Note
In the first sample case, Borya can visit all doctors on days 2, 3 and 4.In the second sample case, Borya can visit all doctors on days 10 and 11.
算出看完所有医生所花的时间，其实就是算出第几天看最后一名医生
#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
int main(){
int n;
cin>>n;
int ans=0;
while(n--){
int s,d;
cin>>s>>d;
while(s<=ans){
s+=d;
}
ans=s;
}
cout<<ans<<endl;
return 0;
}B. Table Tennis
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
n people are standing in a line to play table tennis. At first, the first two players in the line play a game. Then the loser goes to the end of the line, and the winner plays with the next person from the line, and so on. They play until someone wins k games in a row. This player becomes the winner.For each of the participants, you know the power to play table tennis, and for all players these values are different. In a game the player with greater power always wins. Determine who will be the winner.
Input
The first line contains two integers: n and k (2 ≤ n ≤ 500, 2 ≤ k ≤ 1012) — the number of people and the number of wins after which a player leaves, respectively.The second line contains n integers a1, a2, …, an (1 ≤ ai ≤ n) — powers of the player. It’s guaranteed that this line contains a valid permutation, i.e. all ai are distinct.
Output
Output a single integer — power of the winner.Examples
input
2 2
1 2
output
2
input
4 2
3 1 2 4
output
3
input
6 2
6 5 3 1 2 4
output
6
input
2 10000000000
2 1
output
2
Note
Games in the second sample:3 plays with 1. 3 wins. 1 goes to the end of the line.
3 plays with 2. 3 wins. He wins twice in a row. He becomes the winner.
题目说的是赢k场如果这个人输了就到队列尾部，但如果一次遍历后没人赢的话，赢的肯定就是那个power最多的。这题开始不知道哪被hack了。。后来发现，后一个人赢了前边的没有算进去一次。。。。
#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
int main(){
int n;
LL k;
cin>>n>>k;
int a[505];
for(int i=0;i<n;i++)
cin>>a[i];
LL t=0;
int m=a[0];
for(int i=1;i<n;i++){
if(t>=k){
cout<<m;
return 0;
}
if(m>a[i])
t++;
else
m=a[i],t=1;
}
cout<<m<<endl;
return 0;
}C. Short Program
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
Petya learned a new programming language CALPAS. A program in this language always takes one non-negative integer and returns one non-negative integer as well.In the language, there are only three commands: apply a bitwise operation AND, OR or XOR with a given constant to the current integer. A program can contain an arbitrary sequence of these operations with arbitrary constants from 0 to 1023. When the program is run, all operations are applied (in the given order) to the argument and in the end the result integer is returned.
Petya wrote a program in this language, but it turned out to be too long. Write a program in CALPAS that does the same thing as the Petya’s program, and consists of no more than 5 lines. Your program should return the same integer as Petya’s program for all arguments from 0 to 1023.
Input
The first line contains an integer n (1 ≤ n ≤5·105) — the number of lines.Next n lines contain commands. A command consists of a character that represents the operation (“&”, “|” or “^” for AND, OR or XOR respectively), and the constant xi 0 ≤ xi ≤ 1023.
Output
Output an integer k (0 ≤ k ≤ 5) — the length of your program.Next k lines must contain commands in the same format as in the input.
Examples
input
3
| 3
^ 2
| 1
output
2
| 3
^ 2
input
3
& 1
& 3
& 5
output
1
& 1
input
3
^ 1
^ 2
^ 3
output
0
Note
You can read about bitwise operations in https://en.wikipedia.org/wiki/Bitwise_operation.Second sample:
Let x be an input of the Petya’s program. It’s output is ((x&1)&3)&5 = x&(1&3&5) = x&1. So these two programs always give the same outputs.
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <iostream>
#include <map>
#include <queue>
#include <cstdlib>
#include <cstring>
#include <string>
#include <ctime>
#include <vector>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef double db;
typedef long double ldb;
const ll MOD = 1000000009;
const int INF = 0x3f3f3f3f;
const ll LL_INF = 0x3f3f3f3f3f3f3f3f;
const db PI = acos(-1);
const db ERR = 1e-8;
#define rep(i, n) for(int i=0;i<n;i++)
int v[10];
int main(){
int N,i,j;
scanf("%d",&N);
for (i=0;i<10;i++) v[i]=2;
while(N--){
char c[2];
int t;
scanf("%s %d",c,&t);
if(c[0]=='|'){
for (i=0;i<10;i++)
if(t&(1<<i))
v[i]=1;
}
if(c[0]=='&'){
for (i=0;i<10;i++)
if(!(t&(1<<i)))
v[i] = 0;
}
if(c[0]=='^')
for(i=0;i<10;i++)
if(t&(1<<i))
v[i]^= 1;
}
int v1=0,v2=0,v3=0;
for(i=0;i<10;i++){
if(v[i]==0)
v1|=(1<<i);
if(v[i]==1)
v2|=(1<<i);
if(v[i]==3)
v3|=(1<<i);
}
printf("3\n");
printf("| %d\n",v2);
printf("& %d\n",1023^v1);
printf("^ %d\n",v3);
return 0;
}
