Note In the first sample you give the set of numbers {1,?3,?5} to the first friend and the set of numbers {2} to the second friend. Note that if you give set {1,?3,?5} to the first friend, then we cannot give any of the numbers 1 , 3 , 5 t
Note
In the first sample you give the set of numbers {1,?3,?5} to the first friend and the set of numbers {2} to the second friend. Note that if you give set {1,?3,?5} to the first friend, then we cannot give any of the numbers 1, 3, 5 to the second friend.
In the second sample you give the set of numbers {3} to the first friend, and the set of numbers {1,?2,?4} to the second friend. Thus, the answer to the problem is 4.
二分渣渣把二分又写跪了，总是分不清l与r的关系o(╯□╰)o，我竟然l和r都判断了一下。这题竟然l和r都能过。
这题就是枚举一下中间结果，对a周期为x-1，b周期为y-1，只有当i为y的倍数时，只能让a取，当i为x的倍数时，只
能让b取，算一下x，y的倍数时两个都不能取得，a的总数量减去只能a取的，b的总数量减去只能b取的 ，剩下的要
取的和小于等于两个都能取得，这个值就是有效值。
代码：
<mark style="color:transparent">本文来源gaodaimacom搞#^代%!码网@</mark>#include #include #include using namespace std;long long cnt1,cnt2,x,y;long long gcd(long long a,long long b){ return b==0?a:gcd(b,a%b);}bool judge(long long v){ long long t1,t2,t3; t1=v/x; t2=v/y; t3=v/(x*y/gcd(x,y)); long long temp1,temp2,temp3; temp1=max((long long)0,cnt1-t2+t3);//t2-t3是只能a取的，cnt1-只能a取的，就是剩下a没取的 temp2=max((long long)0,cnt2-t1+t3);//t1-t3是只能b取的，cnt2-只能b取的，就是剩下b没取的 temp3=max((long long)0,v-t1-t2+t3);//x，y都能取的 if(temp3>=temp1+temp2)//a，b都能取的大于要大于等于a，b没取的和 return true; else return false;}int main(){ scanf("%I64d%I64d%I64d%I64d",&cnt1,&cnt2,&x,&y); long long l=1,r=2000000000; long long u=0; while(l>1; if(judge(m)) { u=m; r=m; } else { l=m+1; } } // long long ans; // if(judge(r)) // ans=r; // else // ans=l; printf("%I64d\n",u); return 0;}
转载请注明原文链接：B. Friends and Presents（Codeforces Round #275(div2)
