Sum of Numerators

题目链接
学习资料

题意：给定n,k;每一项ai/2^k约分后， ∑ \sum ∑分子。

假设n = 8，k = 2;
初始分子：
1 2 3 4 5 6 7 8
res = （1+8) * 8 / 2 = 36
第一轮约分后：
1 1 3 2 5 3 7 4
[1-n]之间的偶数都减半
res -= (1+4) * 4/ 2 =26
第二轮约分后：
1 1 3 1 5 3 7 2
[1-n/2]之间的偶数都减半
res -= (1+2) * 2 / 2 = 23

直到!n || !k 截止；

Code:

int T;
int main() {cin >> T;while(T--) {ll n,k;cin>>n>>k;ll sum = (1+n)*n/2;while(n&&k) {n>>=1;k--;sum -= (1+n)*n/2;}cout<<sum<<endl;}return 0;
}

题目变式：
可以每一项都除以3^k,4^k…做法同理

逆水行舟，不进则退~

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Sum of Numerators

Sum of Numerators

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