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https://github.com/Kiritow/OJ-Problems-Source.git
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45 lines
900 B
C++
45 lines
900 B
C++
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#include <bits/stdc++.h>
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using namespace std;
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typedef long long LL;
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int t;
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/**
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如果 n 是 奇数，直接输出 0. (奇数+奇数+偶数 或 偶数+偶数+偶数 ==== 偶数)
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如果是偶数，并且是 2 的某个次方，输出 0.
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否则统计二进制中 1 的个数。
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两个 1 答案为 1
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三个 1 答案为 4
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四个 1 答案为 13
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五个 1 答案为 40
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六个 1 答案为 121
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可以把答案单独开个数组 F[ N ] = F [ N -1 ] * 3 +1;
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求得通项公式为
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a(n)=(3^(n-1)-1)/2
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*/
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inline int getTbyte(LL inc)
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{
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int c=0;
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while(inc>0)
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{
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if(inc&1) ++c;
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inc>>=1;
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}
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return c;
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}
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int main()
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{
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scanf("%d",&t);
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while(t--)
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{
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LL inc;
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scanf("%lld",&inc);
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int n=getTbyte(inc);
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if(inc&1 || n == 1)
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{
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printf("0\n");continue;
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}
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printf("%lld\n",((LL)pow(3,n-1)-1)/2);
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}
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return 0;
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}
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