求s=1+1(1+2)+1(1+2+3)….+1(1+2+3….+n)的值
求s=1+1/(1+2)+1/(1+2+3)….+1/(1+2+3….+n)的值 #include <stdio.h> float fun(int n) { int i,s1=0; float s=0.0; for(i=1;i<=n;i++) {s1=s1+i; /求每一项的分母/ s=s+1.0/s1; /求多项式的值/ } return s; } void main() { int n; float s; void NONO ( ); printf(" Please enter N:"); scanf("%d", &n); s = fun(n); printf("the result is: %f ", s); NONO(); } void NONO ( ) {/ 本函数用于打开文件，输入数据，调用函数，输出数据，关闭文件。 / FILE fp, wf ; int i, n ; float s;
fp = fopen("in.dat","r") ; wf = fopen("out.dat","w") ; for(i = 0 ; i < 10 ; i++) { fscanf(fp, "%d", &n) ; s = fun(n) ; fprintf(wf, "%f ", s) ; } fclose(fp) ; fclose(wf) ; }
求s=1+1/(1+2)+1/(1+2+3)….+1/(1+2+3….+n)的值 #include