Educational Codeforces Round 13 A、B、C、D

A. Johny Likes Numbers
time limit per test
0.5 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

Johny likes numbers n and k very much. Now Johny wants to find the smallest integer x greater than n, so it is divisible by the number k.

Input

The only line contains two integers n and k (1 ≤ n, k ≤ 109).

Output

Print the smallest integer x > n, so it is divisible by the number k.

Examples
input
5 3
output
6
input
25 13
output
26
input
26 13
output
39
题意：找到一个大于n并且被k整除的最小的数；
思路：除+1；
#include<bits/stdc++.h>
using namespace std;
#define ll __int64
#define mod 1000000007
#define pi (4*atan(1.0))
const int N=1e3+10,M=1e6+10,inf=1e9+10;
int main()
{
    int x,y,z,i,t;
    scanf("%d%d",&x,&y);
    printf("%I64d\n",(ll)(x/y+1)*y);
    return 0;
}
代码
B. The Same Calendar
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

The girl Taylor has a beautiful calendar for the year y. In the calendar all days are given with their days of week: Monday, Tuesday, Wednesday, Thursday, Friday, Saturday and Sunday.

The calendar is so beautiful that she wants to know what is the next year after y when the calendar will be exactly the same. Help Taylor to find that year.

Note that leap years has 366 days. The year is leap if it is divisible by 400 or it is divisible by 4, but not by 100(https://en.wikipedia.org/wiki/Leap_year).

Input

The only line contains integer y (1000 ≤ y < 100'000) — the year of the calendar.

Output

Print the only integer y' — the next year after y when the calendar will be the same. Note that you should find the first year after y with the same calendar.

Examples
input
2016
output
2044
input
2000
output
2028
input
50501
output
50507
Note

Today is Monday, the 13th of June, 2016.

 题意：找到这年之后日历相同的年份；

思路：天数整除7并且是同是闰年或平年；

#include<bits/stdc++.h>
using namespace std;
#define ll __int64
#define mod 1000000007
#define pi (4*atan(1.0))
const int N=1e3+10,M=1e6+10,inf=1e9+10;
int check(int x)
{
    if((x%4==0&&x%100!=0)||x%400==0)
    return 366;
    return 365;
}
int main()
{
    int x,y,z,i,t;
    scanf("%d",&x);
    int ans=0;
    for(i=x;;i++)
    {
        ans+=check(i)%7;
        if(ans%7==0&&(check(x)==check(i+1)))
            break;
    }
    cout<<i+1<<endl;
    return 0;
}
代码
C. Joty and Chocolate
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

Little Joty has got a task to do. She has a line of n tiles indexed from 1 to n. She has to paint them in a strange pattern.

An unpainted tile should be painted Red if it's index is divisible by a and an unpainted tile should be painted Blue if it's index is divisible by b. So the tile with the number divisible by a and b can be either painted Red or Blue.

After her painting is done, she will get p chocolates for each tile that is painted Red and q chocolates for each tile that is painted Blue.

Note that she can paint tiles in any order she wants.

Given the required information, find the maximum number of chocolates Joty can get.

Input

The only line contains five integers n, a, b, p and q (1 ≤ n, a, b, p, q ≤ 109).

Output

Print the only integer s — the maximum number of chocolates Joty can get.

Note that the answer can be too large, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type.

Examples
input
5 2 3 12 15
output
39
input
20 2 3 3 5
output
51

 题意：1-n中每个整除a的数可以得到p；1-n中每个整除b的数可以得到q；

　　　　问最多可以得到多少；

思路：简单的容斥；

#include<bits/stdc++.h>
using namespace std;
#define ll __int64
#define mod 1000000007
#define pi (4*atan(1.0))
const int N=1e3+10,M=1e6+10,inf=1e9+10;
ll gcd(ll x,ll y)
{
    return y==0?x:gcd(y,x%y);
}
int main()
{
    int x,y,z,i,t;
    ll n,a,b,p,q;
    scanf("%I64d%I64d%I64d%I64d%I64d",&n,&a,&b,&p,&q);
    ll aa=n/a;
    ll bb=n/b;
    ll ans=aa*p+bb*q;
    ll lcm=a*b/gcd(a,b);
    ans-=n/lcm*min(p,q);
    printf("%I64d\n",ans);
    return 0;
}
代码
D. Iterated Linear Function
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

Consider a linear function f(x) = Ax + B. Let's define g(0)(x) = x and g(n)(x) = f(g(n - 1)(x)) for n > 0. For the given integer values A, B,n and x find the value of g(n)(x) modulo 109 + 7.

Input

The only line contains four integers A, B, n and x (1 ≤ A, B, x ≤ 109, 1 ≤ n ≤ 1018) — the parameters from the problem statement.

Note that the given value n can be too large, so you should use 64-bit integer type to store it. In C++ you can use the long longinteger type and in Java you can use long integer type.

Output

Print the only integer s — the value g(n)(x) modulo 109 + 7.

Examples
input
3 4 1 1
output
7
input
3 4 2 1
output
25
input
3 4 3 1
output
79

 思路：赤裸裸的矩阵快速幂；

#include<bits/stdc++.h>
using namespace std;
#define ll __int64
#define mod 1000000007
#define pi (4*atan(1.0))
const int N=1e3+10,M=1e6+10,inf=1e9+10;
struct is
{
    ll a[10][10];
};
ll x,m,k,c;
is juzhenmul(is a,is b,ll hang ,ll lie)
{
    int i,t,j;
    is ans;
    memset(ans.a,0,sizeof(ans.a));
    for(i=1;i<=hang;i++)
    for(t=1;t<=lie;t++)
    for(j=1;j<=lie;j++)
    {
        ans.a[i][t]+=(a.a[i][j]*b.a[j][t]);
        ans.a[i][t]%=mod;
    }
    return ans;
}
is quickpow(is ans,is a,ll x)
{
    while(x)
    {
        if(x&1)  ans=juzhenmul(ans,a,2,2);
        a=juzhenmul(a,a,2,2);
        x>>=1;
    }
    return ans;
}
int main()
{
    int y,z,i,t;
    ll a,b,n,x;
    scanf("%I64d%I64d%I64d%I64d",&a,&b,&n,&x);
    is base;
    base.a[1][1]=a;
    base.a[1][2]=0;
    base.a[2][1]=b;
    base.a[2][2]=1;
    is ans;
    memset(ans.a,0,sizeof(ans.a));
    ans.a[1][1]=1;
    ans.a[2][2]=1;
    ans=quickpow(ans,base,n);
    printf("%I64d\n",(x*ans.a[1][1]+ans.a[2][1])%mod);
    return 0;
}
代码

 

 

posted @ 2016-06-21 15:40  jhz033  阅读(190)  评论(0)  编辑  收藏  举报