System.out.println(\
}
public static void main(String args[]) {
Test t = new Test();
long begin = System.currentTimeMillis();
t.BigestNumber(10000000);
long end = System.currentTimeMillis();
System.out.println(\总时间\秒\
}
}
结果：
f(10000000)=7000001
总时间5秒
1、将一整数逆序后放入一数组中（要求递归实现）
void convert(int *result, int n) {
if(n>=10)
convert(result+1, n/10);
*result = n;
}
int main(int argc, char* argv[]) {
int n = 123456789, result[20]={};
convert(result, n);
printf(\
for(int i=0; i<9; i++)
printf(\
}
2、求高于平均分的学生学号及成绩（学号和成绩人工输入）
double find(int total, int n) {
int number, score, average;
scanf(\
if(number != 0) {
scanf(\
average = find(total+score, n+1);
if(score >= average)
printf(\
return average;
} else {
printf(\
return total/n;
}
