p=109+7p=10^9+7p=109+7
(∑i=1naibi) mod p\left( \sum_{i=1}^n \frac{a_i}{b_i}\right)\bmod p(∑i=1nbiai)modp 是否等于 [∑i=1n(aibi mod p)] mod p \left[ \sum_{i=1}^n \left(\frac{a_i}{b_i}\bmod p\right) \right]\bmod p[∑i=1n(biaimodp)]modp?