Codeforces #275 div2_html/css_WEB-ITnose

Link: codeforce #275 div2

A.Counterexample

Question: Given the left and right intervals [l, r], find the three numbers a, b in the interval ,c makes a and b mutually prime,

b and c are mutually prime, but a and c are not mutually prime. If they do not exist, output -1

Analysis: Find continuous even-odd sequences Just

#include<stdio.h>int main(){    __int64 l,r;    scanf("%I64d%I64d",&l,&r);    if(l%2)        l++;    if(r-l <br> <p></p> <p>B.Friends and Presents</p> <p>Question: Construct two sequences, the first sequence is There are cnt1 numbers and there cannot be multiples of x. </p> <p>The second sequence must have cnt2 numbers and there cannot be multiples of y. Two sequences cannot have the same number. </p> <p> is required to find the minimum value of the maximum number in the two sequences. </p> <p>Analysis: m=num - num / x, is the number of numbers from 1 to num that does not contain multiples of x </p> <p> n=num - num / y, is the number from 1 to num that does not The number of numbers containing multiples of ;=cnt2</p> <p>Because two sequences cannot have the same number, so cnt1 cnt2  </p>
<p>Then binary search for the minimum value of num</p> <p></p> <p> </p>
<p> </p> <pre name="code" class="sycode">#include<stdio.h>int main(){    __int64 x,y,cnt1,cnt2,m,n;    __int64 l,r,mid;    scanf("%I64d%I64d%I64d%I64d",&cnt1,&cnt2,&x,&y);    l=1;    r=1e12;    while(l<r mid="(l+r)/2;" m="mid-mid/x;" n="mid-mid/y;" if>=cnt1&&n>=cnt2&&mid-mid/(x*y)>=cnt1+cnt2)            r=mid;        else            l=mid+1;    }    printf("%I64d\n",r);    return 0;}</r></stdio.h>

C.Diverse Permutation

Question: Find a sequence containing n numbers from 1-n, The number of different absolute values ​​of the difference between two adjacent elements is required to be k

Analysis: There are n-1 differences in n numbers, and it is necessary to ensure that the absolute values ​​of the k differences are different,

Then there are n-k-1 differences with the same difference. You can first output the n-k numbers between [1, n-k] in order,

and then output the minimum and maximum values ​​of the remaining numbers in sequence until you lose Until n numbers are reached.

#include<stdio.h>int main(){    int n,k,i,j,num;    scanf("%d%d",&n,&k);    num=n-k-1;    for(i=1;i <br> <br> <p></p></stdio.h>