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### 题目分析
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* 一年中有$52$个星期, $7 * 52 = 364$
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* 星期一：$x$元
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* 星期二:$x+k$
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* 星期三:$x+2k$
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* 星期四:$x+3k$
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* 星期五:$x+4k$
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* 星期六:$x+5k$
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* 星期日:$x+6k$
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* 目标是筹集$n$元
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问题：$x$和$k$是怎么样来取值的？输出 $x$ **尽可能大**，$k$ **尽可能小的**。
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$$1<=x<=100,k>=1$$
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思考：
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$$n=52\times (x+ (x+k) +(x+2k)+(x+3k)+(x+4k)+(x+5k)+(x+6k)) \\
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\Leftrightarrow \\
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n=52 \times (7x+21k)
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$$
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```c++
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for(int x=100;x>=1;x--){
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for(int k=1;52 * (7 * x + 21 * k) <= n;k++){
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if(n==52*(7*x+21*k))
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{
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cout<<x<<endl;
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cout<<k<<endl;
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exit(0);
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}
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}
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}
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```
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已知$x>=1,y>=1$，且$x+y=5$
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问$x=?,y=?$
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显然：$x<=4,y<=4$
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$1<=x<=4,1<=y<=4$
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(1)x=1,y=4
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(2)x=2,y=3
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(3)x=3,y=2
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(4)x=4,y=1
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