测试数据 #0: Accepted, time = 0 ms, mem = 0 KiB, score = 10
测试数据 #1: Accepted, time = 0 ms, mem = 0 KiB, score = 10
测试数据 #2: Accepted, time = 0 ms, mem = 0 KiB, score = 10
测试数据 #3: Accepted, time = 0 ms, mem = 0 KiB, score = 10
测试数据 #4: Accepted, time = 0 ms, mem = 0 KiB, score = 10
测试数据 #5: Accepted, time = 0 ms, mem = 0 KiB, score = 10
测试数据 #6: Accepted, time = 0 ms, mem = 0 KiB, score = 10
测试数据 #7: Accepted, time = 0 ms, mem = 0 KiB, score = 10
测试数据 #8: Accepted, time = 0 ms, mem = 0 KiB, score = 10
测试数据 #9: Accepted, time = 0 ms, mem = 0 KiB, score = 10
Accepted, time = 0 ms, mem = 0 KiB, score = 100

我觉得很难......

PS:我是菜鸟

#终于AC了

P1308埃及分数Accepted
记录信息
评测状态 Accepted
题目 P1308 埃及分数
递交时间 2015-10-17 21:39:08
代码语言 C++
评测机 VijosEx
消耗时间 730 ms
消耗内存 528 KiB
评测时间 2015-10-17 21:39:10
评测结果
编译成功

foo.cpp: In function 'void print()':
foo.cpp:16:23: warning: unknown conversion type character 'l' in format [-Wformat=]
printf("%lld ",ans[i]);
^
foo.cpp:16:23: warning: too many arguments for format [-Wformat-extra-args]
foo.cpp: In function 'int main()':
foo.cpp:59:28: warning: unknown conversion type character 'l' in format [-Wformat=]
scanf("%lld %lld",&a,&b);
^
foo.cpp:59:28: warning: unknown conversion type character 'l' in format [-Wformat=]
foo.cpp:59:28: warning: too many arguments for format [-Wformat-extra-args]
测试数据 #0: Accepted, time = 0 ms, mem = 528 KiB, score = 10
测试数据 #1: Accepted, time = 15 ms, mem = 528 KiB, score = 10
测试数据 #2: Accepted, time = 15 ms, mem = 524 KiB, score = 10
测试数据 #3: Accepted, time = 0 ms, mem = 528 KiB, score = 10
测试数据 #4: Accepted, time = 0 ms, mem = 524 KiB, score = 10
测试数据 #5: Accepted, time = 15 ms, mem = 524 KiB, score = 10
测试数据 #6: Accepted, time = 171 ms, mem = 528 KiB, score = 10
测试数据 #7: Accepted, time = 15 ms, mem = 524 KiB, score = 10
测试数据 #8: Accepted, time = 15 ms, mem = 524 KiB, score = 10
测试数据 #9: Accepted, time = 484 ms, mem = 528 KiB, score = 10
Accepted, time = 730 ms, mem = 528 KiB, score = 100
代码

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;

long long ans[21],now[21];
int dep;
long long GCD(long long x,long long y){return y?GCD(y,x%y):x;}
void print()
{
for(int i=1;ans[i];i++)
printf("%lld ",ans[i]);
}
bool search(long long a,long long b,int k)
{
if(k==dep){
if(a==1){
now[k]=b;
if(now[k]<ans[k]||ans[k]==0)
memcpy(ans,now,sizeof(now));
return true;
}
return false;
}
bool flag=false;
long long first=b%a?b/a+1:b/a;
for(long long i=max(now[k-1]+1,first);;i++)
{
if(b*(dep+1-k)<=i*a) break;
now[k]=i;
long long x=a*i-b;
long long y=b*i;
long long kt=GCD(x,y);
if(search((long long)x/kt,(long long)y/kt,k+1))
flag=true;
}
return flag;
}
void work(int a,int b)
{
for(dep=1;;dep++)
{
if(search((long long)a,(long long)b,1)){
print();
break;
}
}
}
int main()
{
long long a,b;
scanf("%lld %lld",&a,&b);
int kt=GCD(a,b);
work(a/kt,b/kt);
return 0;
}

话说此题真的非常经典，但是确实X

真的.....看我无剪枝暴搜程序过全点，真的很滑稽....

没什么好说的，就是纯暴力搜索 。。。。最外层枚举分数数量搜索 。。。。

贴个主体部分

procedure search(z:longint;x,y:int64);

var i:longint;

q,t,p,j,k:int64;

begin

if (y mod x=0) and (y div x>a[z-1]) then

begin

a[z]:=y div x;

if not f or (a[z]0) and (z