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?????????????????????? ? ? ?????? The mook jong

Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 767????Accepted Submission(s): 522

Problem Description
ZJiaQ want to become a strong man, so he decided to play the mook jong。ZJiaQ want to put some mook jongs in his backyard. His backyard ?consist of n bricks that is 1*1,so it is 1*n。ZJiaQ want to put a mook ? ? ? ?jong in a brick. because of the hands of the mook jong， the distance of two mook jongs should be equal or more than 2 bricks. Now ZJiaQ want to know how many ways can ZJiaQ put mook jongs legally(at least one mook jong).

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Input

There ar multiply cases. For ?each case, there is a single integer n( 1 < = n < = 60)

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Output

Print the ways in a single line for each case.

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Sample Input

1 2 3 4 5 6

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Sample Output

1 2 3 5 8 12

?Source

BestCoder Round #50 (div.2)

題意

大致模型就是說，有一個1*N的格子，（就像下面這樣），往里面塞東西，但是每兩個之間距離大于等于2.

1

2

3

···

···

n

對于樣例，輸入4有如下5種解法

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? ??

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通過自己舉幾個例子發現當輸入從1~60的時候

輸出是這樣的??????????????????? 1 2 3 5 8 12······

是不是好像發現了什么？

再結合每兩個元素之間至少要2個間距，仔細思考之后得出公式?????

a[i]=a[i-1]+a[i-3]+1;

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Tips

當N接近60的時候數據會很大，注意數據的存儲類型！

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#include<iostream> using?namespace?std; long?long?a[61]={0,1,2,3}; int?main(){int?n;for(int?i=4;i<=60;i++){a[i]=a[i-1]+a[i-3]+1;}while(cin>>n){cout<<a[n]<<endl;}return?0; }

轉載于:https://my.oschina.net/zmixed/blog/625406

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