-sharpinclude <stdio.h>
int x = 10000000000;
int n = 0;
int main() {
for(int i = 0; i<x; iPP) n = n+i;
printf("%d" ,n);
printf("%s" ,": ok");
}then replace the statements of x n and I with double
but the final result may deviate from the real result
1x 2m 3+. + 10000000000, the result of this formula is 5.0000000005Eq19, which is too large to be out of the range of int.
if you only measure the time of addition, it is recommended to use loop nesting, like this
for (int i = 0; i < 1000000; iPP) {
n = 0;
for (int j = 0; j < 10000; jPP) {
n += j;
}
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