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And at the same time within 1000 can be divided exactly by 3, 5, 7

Time：10-28

A nested loop what is the problem of

CodePudding user response:

Int I=3, j=5, k=7, n.
For (n=I * j * k; n<=1000; N +=I * j * k)
Cout CodePudding user response:

reference 1st floor to hit dead Newton's apple's reply:

int I=3, j=5, k=7, n.
For (n=I * j * k; n<=1000; N +=I * j * k)
Cout

Simple and efficient a look is a great god

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