题目非常简单， 答案就是 n/n+n/(n-1)+...+n/1;
要求分数输出。套用分数模板。。
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
const int maxn = 500; struct fraction {
long long numerator; // 分子
long long denominator; // 分母
fraction() {
numerator = 0;
denominator = 1;
}
fraction(long long num) {
numerator = num;
denominator = 1;
}
fraction(long long a, long long b) {
numerator = a;
denominator = b;
this->reduction();
} void operator = (const long long num) {
numerator = num;
denominator = 1;
this->reduction();
} void operator = (const fraction &b) {
numerator = b.numerator;
denominator = b.denominator;
this->reduction();
} fraction operator + (const fraction &b) const {
long long gcdnum = __gcd(denominator, b.denominator);
fraction tmp = fraction(numerator*(b.denominator/gcdnum) + b.numerator*(denominator/gcdnum), denominator/gcdnum*b.denominator);
tmp.reduction();
return tmp;
} fraction operator + (const int &b) const {
return ((*this) + fraction(b));
} fraction operator - (const fraction &b) const {
return ((*this) + fraction(-b.numerator, b.denominator));
} fraction operator - (const int &b) const {
return ((*this) - fraction(b));
} fraction operator * (const fraction &b) const {
fraction tmp = fraction(numerator*b.numerator, denominator * b.denominator);
tmp.reduction();
return tmp;
} fraction operator * (const int &b) const {
return ((*this) * fraction(b));
} fraction operator / (const fraction &b) const {
return ((*this) * fraction(b.denominator, b.numerator));
} void reduction() {
if (numerator == 0) {
denominator = 1;
return;
}
long long gcdnum = __gcd(numerator, denominator);
numerator /= gcdnum;
denominator /= gcdnum;
}
void print() {
if (denominator == 1) printf("%lld\n", numerator);
else {
long long num = numerator/denominator;
long long tmp = num;
int len = 0;
while (tmp) {
len++;
tmp/=10;
} for (int i = 0; i < len; i++) printf(" ");
if (len != 0) printf(" ");
printf("%lld\n",numerator%denominator); if (num != 0) printf("%lld ", num);
tmp = denominator;
while (tmp) {
printf("-");
tmp/=10;
}
puts(""); for (int i = 0; i < len; i++) printf(" ");
if (len != 0) printf(" ");
printf("%lld\n",denominator);
}
}
} f[maxn]; int main() {
int n;
while (scanf("%d", &n) != EOF) { f[0] = 0;
for (int i = 1; i <= n; i++)
f[i] = f[i-1] + fraction(1, i);
f[n] = f[n] * n; f[n].print();
}
return 0;
}