给出一个真分数a/b,使用单位分数的和(1/a,a是自然数),拆分的少的比拆分的多的好 拆分的分数相同情况下 ,最小的分数越大越好
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
int a, b, maxd;
typedef long long LL;
LL gcd(LL a, LL b) {
return b == 0 ? a : gcd(b, a%b);
}
// 返回满足1/c <= a/b的最小c
inline int get_first(LL a, LL b) {
return b/a+1;
}
const int maxn = 100 + 5;
LL v[maxn], ans[maxn];
// 如果当前解v比目前最优解ans更优,更新ans
bool better(int d) {
for(int i = d; i >= 0; i--) if(v[i] != ans[i]) {
return ans[i] == -1 || v[i] < ans[i];
}
return false;
}
// 当前深度为d,分母不能小于from,分数之和恰好为aa/bb
bool dfs(int d, int from, LL aa, LL bb) {
if(d == maxd) {
if(bb % aa) return false; // aa/bb必须是埃及分数
v[d] = bb/aa;
if(better(d)) memcpy(ans, v, sizeof(LL) * (d+1));
return true;
}
bool ok = false;
from = max(from, get_first(aa, bb)); // 该层枚举的起点
for(int i = from; ; i++) {
// 剪枝:如果剩下的maxd+1-d个分数全部都是1/i,加起来仍然不超过aa/bb,则无解
if(bb * (maxd+1-d) <= i * aa) break;
v[d] = i;
// 计算aa/bb - 1/i,设结果为a2/b2
LL b2 = bb*i;
LL a2 = aa*i - bb;
LL g = gcd(a2, b2); // 以便约分
if(dfs(d+1, i+1, a2/g, b2/g)) ok = true;
}
return ok;
}
int main() {
int kase = 0;
while(cin >> a >> b) {
int ok = 0;
for(maxd = 1; maxd <= 100; maxd++) {
//从小到大枚举深度上限maxd,每次执行只考虑深度不超过maxd的结点
memset(ans, -1, sizeof(ans));
if(dfs(0, get_first(a, b), a, b)) { ok = 1; break; }
}
cout << "Case " << ++kase << ": ";
if(ok) {
cout << a << "/" << b << "=";
for(int i = 0; i < maxd; i++) cout << "1/" << ans[i] << "+";
cout << "1/" << ans[maxd] << "\n";
} else cout << "No solution.\n";
}
return 0;
}