Balls Rearrangement

題目連接：

http://acm.hdu.edu.cn/showproblem.php?pid=4611

Description

　　Bob has N balls and A boxes. He numbers the balls from 0 to N-1, and numbers the boxes from 0 to A-1. To find the balls easily, he puts the ball numbered x into the box numbered a if x = a mod A. Some day Bob buys B new boxes, and he wants to rearrange the balls from the old boxes to the new boxes. The new boxes are numbered from 0 to B-1. After the rearrangement, the ball numbered x should be in the box number b if x = b mod B.
　　This work may be very boring, so he wants to know the cost before the rearrangement. If he moves a ball from the old box numbered a to the new box numbered b, the cost he considered would be |a-b|. The total cost is the sum of the cost to move every ball, and it is what Bob is interested in now.
　　

Input

The first line of the input is an integer T, the number of test cases.(0<T<=50)
　　Then T test case followed. The only line of each test case are three integers N, A and B.(1<=N<=1000000000, 1<=A,B<=100000).
　　

Output

For each test case, output the total cost.

Sample Input

3
1000000000 1 1
8 2 4
11 5 3

Sample Output

0
8
16

Hint

題意

有n個數(shù)，給你a和b

T1[i]=i%a，T2[i]=i%b

求sigma(T1[i]-T2[i])

題解：

簡單思考一下，他們之間的差一開始都是0的，當(dāng)遇到某個數(shù)是a的倍數(shù)的時候，他們之間每個數(shù)的差就會增加a，遇到某個數(shù)是b的倍數(shù)的時候，差就會減少b

根據(jù)這個去跑一個lcm周期的答案就好了

然后再暴力算最后一個n%lcm的答案就好了

代碼

#include<bits/stdc++.h> using namespace std;long long gcd(long long a,long long b) {if(b==0)return a;return gcd(b,a%b); } long long lcm(long long a,long long b) {return a*b/gcd(a,b); } vector<long long>p; void init() {p.clear(); } void solve() {init();long long n,a,b;cin>>n>>a>>b;long long c = lcm(a,b);for(long long i=1;i*a<=c;p.push_back(i*a),i++);for(long long i=1;i*b<=c;p.push_back(i*b),i++);p.push_back(0);sort(p.begin(),p.end());p.erase(unique(p.begin(),p.end()),p.end());long long ans = 0,now = 0;for(int i=1;i<p.size()-1;i++){if(p[i]%a==0)now-=a;if(p[i]%b==0)now+=b;ans+=abs(now)*(p[i+1]-p[i]);}long long k = n/c;long long Ans = 0;Ans = Ans + k*ans;k = n%c;int kkk = 0;for(int i=0;i<p.size()-1&&p[i+1]<k;i++){kkk = i+1;if(p[i]%a==0)now-=a;if(p[i]%b==0)now+=b;Ans+=abs(now)*(p[i+1]-p[i]);}if(p[kkk]%a==0)now-=a;if(p[kkk]%b==0)now+=b;Ans+=abs(now)*(k-p[kkk]);cout<<Ans<<endl; } int main() {int t;scanf("%d",&t);while(t--)solve();return 0; }

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