暴力枚举

ZOJ1095

这题只要枚举出2^i * 3^j *5^k *7^l 即可，最后排序

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
const long long N= 2000000000;
void get(int n)
{
if(n%100==11||n%100==12||n%100==13)printf("th");
else if(n%10==1)printf("st");
else if(n%10==2)printf("nd");
else if(n%10==3)printf("rd");
else printf("th");
}
int main()
{
long long i,j,k,l,m,p=1;
long long a[6000];
for(i=1;i<N;i*=2)
{
for(j=1;j<N;j*=3)
{
if(i*j>N)break;
for(k=1;k<N;k*=5)
{
if(i*j*k>N)break;
for(l=1;l<N;l*=7)
{
if((m=i*j*k*l)>N)break;
a[p++]=m;
}
}
}
}
sort(a+1,a+5843);
int n;
while(~scanf("%d",&n)&&n){
printf("The %d",n);
get(n);
printf(" humble number is %d.\n",a
);
}
return 0;
}

ZOJ1110

ZOJ1110简单题

暴力枚举 写好约束条件集合

sp>=pu+s;

pu>=ye+p;

sp>=ye+p;

推出：

0<=ye<=(12+j-max(s+p,y)-p)/2;

ye+p<=pu<=sp-s;

max(s+p,y)<=sp<=12+j;

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#define N 100
using namespace std;
int ans;
int s,p,y,m;
void fun()
{
int sp,pu,ye;
ye=ans;
for(;ye>=0;ye--)
{
for(sp=max(s+p,y)+ye;sp<=12+m;sp++)
{
for(pu=p+ye;pu<=sp-s;pu++)
{
if(pu+ye+sp==12+m)
{
printf("%d %d %d\n",sp,pu,ye);
return;
}
}
}
}
}
int main()
{
while(cin>>s>>p>>y>>m)
{
ans=(12+m-max(s+p,y)-p)/2;//yertle<=ans
fun();
}
return 0;
}

1421: Repeat Number

时间限制: 1 Sec 内存限制: 32 MB

提交: 63 解决: 26

[提交][状态][讨论版]

题目描述

Definition: a+b = c, if all the digits of c are same ( c is more than ten)，then we call a and b are Repeat Number. My question is How many Repeat
Numbers in [x,y].

输入

There are several test cases.

Each test cases contains two integers x, y(1<=x<=y<=1,000,000) described above.

Proceed to the end of file.

输出

For each test output the number of couple of Repeat Number in one line.

样例输入

1 10
10 12

样例输出

5
2

提示

If a equals b, we can call a, b are Repeat Numbers too, and a is the Repeat Numbers for itself.

先打表出所有的repeat number再暴力 出x，y的组合个数

#include<iostream>
#include<cstdio>
using namespace std;
//  2*x<=a+b<=2*y
int a[100],t;
void inti(){
int i,j;
t=0;
for(i=11;i<=11111111;i=i*10+1)
{
for(j=1;j<=9;j++)a[t++]=i*j;
}
//  for(i=0;i<t;i++)printf("%d\n",a[i]);
}
int main(){
inti();
int x,y,p,q,i;
while(cin>>x>>y)
{
for(i=0;i<t;i++)
{
if(a[i]>=2*x){
p=i;
break;
}
}
if(i==t){
printf("0\n");
continue;
}
for(i=t-1;i>=0;i--){
if(a[i]<=2*y)
{
q=i;
break;
}
}
if(i==-1)
{
printf("0\n");
continue;
}
int sum=0;
for(i=p;i<=q;i++)
{
if(a[i]%2==0)// x=[x,a[i]/2] y=[a[i]/2,y]  st:x=6,y=12
sum+=min(a[i]/2-x+1,y-a[i]/2+1);
else  //x=[x,a[i]/2] y=[a[i]/2+1,y] st: x=3,y=10;
sum+=min(a[i]/2-x+1,y-a[i]/2);
}
printf("%d\n",sum);
}

return 0;
}