前n项求和f(n)1121314...1nf(n) 1 \frac{1}{2} \frac{1}{3} \frac{1}{4} ... \frac{1}{n}f(n)121​31​41​...n1​intmain(){intn;inti;doublesum0.0;scanf(%d,n);for(i1;in;i){sum1.0/i;}printf(%f\n,sum);return0;}f(n)1−1213−14...1nf(n) 1 - \frac{1}{2} \frac{1}{3} - \frac{1}{4} ... \frac{1}{n}f(n)1−21​31​−41​...n1​intmain(){intn;inti;doublesum0.0;intsign1;scanf(%d,n);for(i1;in;i){sumsign*1.0/i;sign-sign;}printf(交替求和结果是: %f\n,sum);return0;}