原文链接:/ezoiHQM/article/details/82961266?utm_medium=distribute.pc_relevant.none-task-blog-BlogCommendFromMachineLearnPai2-1.nonecase&depth_1-utm_source=distribute.pc_relevant.none-task-blog-BlogCommendFromMachineLearnPai2-1.nonecase
for (int i=1; i<=1000; ++i) {for (int j=1; i*j<=1000000; ++j) {
所以对于这串代码来说，当i=1,sum[j]=1+2+3+….N;当i=2,sum[j]=2+4+6+8+…..—>N/2;当i=3,sum[j]=3+6+9+12+…..->N/3….—->当i=N,sum[j]=1;
累加得总和为 N+N/2+N/3+N/4+N/5+…….N/N==N*(1+1/2+1/3+1/4+1/5+/1/6+1/7+…1/N)，里面是个调和级数,复杂度为logN,和起来是N*logN的复杂度
