do{
for(i=1;sum-(int)sum>1.0e-6;i+=2){
sum+=qgcf(x,i);
if(sum-(int)sum>1.0e-6){
cout<<“sum=”<<sum<<endl;
}
}
}while(sum-(int)sum>1.0e-6);
本人写的比较混论
4)
for(i=1;sum-(int)sum>1.0e-6;i+=2){
sum+=qgcf(x,i);
if(sum-(int)sum>1.0e-6){
cout<<“sum=”<<sum<<endl;
}
}
}while(sum-(int)sum>1.0e-6);
本人写的比较混论
4)
解决方案
10
10
5
printf(“%.5f”,fa);
5
%.15lg
20
#include <stdio.h>
int main() {
double d;
int i;
d=1.23456789012345;
for (i=1;i<16;i++) {
printf("%.*lg\n",i,d);
}
d=1234567890.12345;
for (i=1;i<16;i++) {
printf("%.*lg\n",i,d);
}
return 0;
}
//1
//1.2
//1.23
//1.235
//1.2346
//1.23457
//1.234568
//1.2345679
//1.23456789
//1.23456789
//1.2345678901
//1.23456789012
//1.234567890123
//1.2345678901235
//1.23456789012345
//1e+009
//1.2e+009
//1.23e+009
//1.235e+009
//1.2346e+009
//1.23457e+009
//1.234568e+009
//1.2345679e+009
//1.23456789e+009
//1234567890
//1234567890.1
//1234567890.12
//1234567890.123
//1234567890.1235
//1234567890.12345
//
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