Codeforces 102202D-A Plus Equals B【思维】难度：**

题意：

A + B is a problem used to test one’s basic knowledge for competitive programming. Here is yet another boring variation of it.

You have two integers, A and B. You want to make them equal. To do so, you can perform several steps, where each step is one of the following:

A+=A
A+=B
B+=A
B+=B
Unfortunately, A + B is a hard problem for us, so you are allowed to make at most 5000 steps.

Input
Two integers A, B are given. (1 ≤ A, B ≤ 10¹⁸).

Output
In the first line, print a single integer n (0 ≤ n ≤ 5000) denoting the number of steps.

In next n lines, print one of the following strings to denote your desired operation: “A+=A”, “A+=B”, “B+=A”, or “B+=B”.

Any sequence of steps that yields the desired result will be judged correct.

Example
inputCopy
2 3
outputCopy
4
B+=B
B+=A
A+=A
A+=A

题解：

虽然这题的数字是在向上叠加的，但我们不能真的去向上叠加，一是因为向上坑太大，没办法处理两个数的关系，二是对于大数会爆long long。观察后可以发现A+=A的本质是B-=B,B+=B的本质是A-=A，这样处理可以保持两个数时刻处于纠缠的状态。因此我们只需遇到偶数就除以二并时刻进行辗转相减即可。

代码：

#include<stdio.h>
int main()
{long long a,b,c,d;int times=0;scanf("%lld%lld",&a,&b);c=a;d=b;while(a!=b){while(a%2==0){times++;a/=2;}while(b%2==0){times++;b/=2;}if(a==b)break;if(a<b)b+=a;else a+=b;times++;}printf("%d\n",times);a=c;b=d;while(a!=b){while(a%2==0){printf("B+=B\n");a/=2;}while(b%2==0){printf("A+=A\n");b/=2;}if(a==b)break;if(a<b){b+=a;printf("B+=A\n");}else{a+=b;printf("A+=B\n");}}
}

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Codeforces 102202D-A Plus Equals B【思维】难度：**

题意：

题解：

代码：

Codeforces 102202D-A Plus Equals B【思维】难度：**相关推荐

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Codeforces 102202D-A Plus Equals B【思维】 难度：**

题意：

题解：

代码：

Codeforces 102202D-A Plus Equals B【思维】 难度：**相关推荐

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Codeforces 102202D-A Plus Equals B【思维】难度：**

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