Problem J
Bits
Input: Standard Input
Output: Standard Output
A bit is a binary digit, taking a logical value of either "1" or "0" (also referred to as "true" or "false" respectively). And every decimal number has a binary representation which is actually a series of bits. If a bit of a number is “1” and it's next bit is also “1” then we can say that the number has a 1 adjacent bit. And you have to find out how many times this scenario occurs for all numbers up to N.
Examples:
Number Binary Adjacent Bits
12 1100 1
15 1111 3
27 11011 2
Input
For each test case, you are given an integer number (0 <= N <= ((2^63)-2)), as described in the statement. The last test case is followed by a negative integer in a line by itself, denoting the end of input file.
Output
For every test case, print a line of the form “Case X: Y”, where X is the serial of output (starting from 1) and Y is the cumulative summation of all adjacent bits from 0 to N.
题意:
给出定义 A(x) 为 x 的二进制表示中 11 的个数.
请你求出 A(x) 的前缀和 S(x)
思路:
如果真正理解了 uva 11038, 那么这道题应该差不多...
分段考虑的话,就是当第 i 与 i-1 位 为1 的时候,
左边 有.... 右边有......
根据乘法原理, 直接乘起来就ok.
再根据加法原理,直接加起来就是答案.
由于答案比较大.所以采用两个数字压位的做法..(学到的好方法.)
我的程序写得非常啰嗦.....不如再去看看别人的.
1 #include<cstdlib> 2 #include<cstdio> 3 #include<iostream> 4 #include<cstring> 5 using namespace std; 6 const long long lim = (long long)1e13; 7 typedef long long BIG[2]; 8 BIG ans; 9 long long n,x1,x2,x3,L; 10 int kase; 11 void add(BIG &a,long long b){ 12 a[1] += b; a[0] += a[1] / lim; a[1] %= lim; 13 } 14 int digit(long long x,int pos){ return (x & (1LL << pos)) != 0; } 15 long long ext(long long n,long long st,long long ed){ 16 if(st > ed) return 0; 17 n >>= st; 18 return n & ((1LL << (ed - st + 1)) - 1); 19 } 20 int main() 21 { 22 freopen("bits.in","r",stdin); 23 freopen("bits.out","w",stdout); 24 while(scanf("%lld",&n), n >= 0){ 25 ans[0] = ans[1] = 0; 26 for(L = 63; L > 0 && !digit(n,L); L--); 27 int a = digit(n,L), b = digit(n,L-1); 28 if(a == b && a == 1 && n != 3) add(ans, ext(n,0,L-2) + 1); 29 a = digit(n,1), b = digit(n,0); 30 add(ans, ext(n,2,L) + (a && b)); 31 for(int i = L-1; i >= 2; --i){ 32 a = digit(n,i), b = digit(n,i-1); 33 x1 = ext(n,i+1,L), x2 = ext(n,0,i-2) + 1, x3 = (1LL << (i-1)); 34 add(ans,(x1 && a && b ? 1 : 0) * x2 + x1 * x3); 35 } 36 add(ans,0); 37 printf("Case %d: ",++kase); 38 if(ans[0]){ 39 printf("%lld",ans[0]); 40 printf("%013lld ",ans[1]); 41 } 42 else printf("%lld ",ans[1]); 43 } 44 return 0; 45 }