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求1/(1/10+1/11+1/12+1/13........+1/19)的整数部分

整数部分是 0

1/12*13+1/13*14+.+1/19*20 =(1/12-1/13)+(1/13-1/14)+。。。+(1/19-1/20) =1/12-1/20 =(5-3)/60 =1/30

1/10×11+1/11×12+1/12×13+...+1/20×21 =(1/10-1/11)+(1/11-1/12)+(1/12-1/13)+...+(1/20-1/21) =1/10-1/11+1/11-1/12+1/12-1/13+...+1/20-1/21 =1/10-1/21 =11/210 裂项相消法。

#include "stdio.h" #include "conio.h" void main() { int i,k=1; float sum=0,sum1; for(i=1;i

#include int main(void) { float sum=0; int n,i; scanf("%d",&n); for(int i=1;i

#include void main() { int n,i; double result,flag;//flag应设为浮点型,1/2会等于0 printf("请输入数值n:"); scanf("%d",&n...

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