Description
[n≥3] ,定义使A3*A4*…*Ak为整数的k [k≥3]叫做兑数。求取区间[3,x]内所有兑数的和加上4是多少。
Input
输入包含多组数据，每一行是一个数x，x<=2^31。
Output
每组测试数据输出对应的一行答案，每组数据间有两个空行。具体输出格式见样例。
Sample Input
5
10
Sample Output
Case # 1: 8 -ti - hen- shui -
Case # 2: 16 -ti - hen- shui -
Source
Unknown
思路：这世道什么鬼题，我一开始发现了规律：如果x是2的幂次，结果为2*x；否则为x的前一个2的幂次n，结果为2*n。然后很迷茫该怎么找2的幂次才能解决，后来想到了暴力打表：直接输出结果。但还是wa。妈耶，到后来才发现，还需要添加几个很大的数....
#include <cstdio>
#include <iostream>
#include <
string>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <map>
#include <vector>
using namespace std;
#define ll long long
ll x, sum, f[100+
8];
int id =
0;
bool rule(
int a)
{
return (a&(a-
1))?
false:
true;
}
void init()
{
f[0] =
4;
f[1] =
8;
f[2] =
16;
f[3] =
32;
f[4] =
64;
f[5] =
128;
f[6] =
256;
f[7] =
512;
f[8] =
1024;
f[9] =
2048;
f[10] =
4096;
f[11] =
8192;
f[12] =
16384;
f[13] =
32768;
f[14] =
65536;
f[15] =
131072;
f[16] =
262144;
f[17] =
524288;
f[18] =
1048576;
f[19] =
2097152;
f[20] =
4194304;
f[21] =
8388608;
f[22] =
16777216;
f[23] =
33554432;
f[24] =
67108864;
f[25] =
134217728;
f[26] =
268435456;
f[27] =
536870912;
f[28] =
1073741824;
f[29] =
2147483648;
f[30] =
4294967296;
}
int main()
{
init();
int ii =
1;
while(~scanf(
"%lld", &
x))
{
sum =
0;
if(x ==
3)
{
printf("Case # %d: 4 -ti - hen- shui -\n\n\n", ii);
ii++
;
continue;
}
ll sign;
if(rule(x))sum = x*
2;
else
{
for(
int i =
1; i<
31; i++
)
{
// cout<<f[i]<<"------"<<endl;
if(f[i]>
x)
{
sign = f[i-
1];
// cout<<sign<<endl;
break;
}
}
sum = sign*
2;
}
printf("Case # %d: %lld -ti - hen- shui -\n\n\n", ii, sum);
ii++
;
}
return 0;
}
转载于:https://www.cnblogs.com/RootVount/p/11005438.html
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