Codeforces--598A--Tricky Sum(数学)
Tricky Sum
Tricky SumCrawling in process... Crawling failed Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u
Description
In this problem you are to calculate the sum of all integers from 1 to n, but you should take all powers of two with minus in the sum.
For example, for n = 4 the sum is equal to - 1 - 2 + 3 - 4 = - 4, because 1, 2 and 4 are 20, 21 and 22 respectively.
Calculate the answer for t values of n.
Input
The first line of the input contains a single integer t (1 ≤ t ≤ 100) — the number of values of n to be processed.
Each of next t lines contains a single integer n (1 ≤ n ≤ 109).
Output
Print the requested sum for each of t integers n given in the input.
Sample Input
2 4 1000000000
-4 499999998352516354
Sample Output
Hint
The answer for the first sample is explained in the statement.
1--n全部加起来,如果是2的次幂 数字就为负的,求最后的和,先利用等差序列公式(n+1)*n/2,然后扣除2的次幂
#include<cstdio> #include<iostream> #include<cstring> #include<cmath> #include<algorithm> using namespace std; int main() { int t; cin>>t; while(t--) { __int64 n; cin>>n; __int64 ans=(n+1)*n/2-2; for(int i=1;pow(2,i)<=n;i++) ans-=pow(2,i+1); cout<<ans<<endl; } return 0; }