調和級數近似求和公式推導(轉載)

原文鏈接: https://blog.csdn.net/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

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for (int i=1; i<=1000; ++i) {
        for (int j=1; i*j<=1000000; ++j) {

所以對於這串代碼來說，當i=1,sum[j]=1+2+3+….N;當i=2,sum[j]=2+4+6+8+…..—>N/2;當i=3,sum[j]=3+6+9+12+…..->N/3….—->當i=N,sum[j]=1;

累加得總和爲 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)，裏面是個調和級數,複雜度爲logN,和起來是N*logN的複雜度