第六届湘潭市程序设计竞赛 -Happy Number
时间：2014-06-08 05:44:33
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| Accepted : 110 | Submit : 263 | |
| Time Limit : 1000 MS | Memory Limit : 65536 KB | |
Problem DescriptionRecently, Mr. Xie learn the concept of happy number. A happy number is a number contain all digit 7 or only 1 digit other than 7. For example, 777 is a happy number because 777 contail all digit 7, 7177 and 87777 both happy number because only 1 digit other than 7. Whereas 887,799 9807,12345, all of them are not happy number. Now Mr. xie want to know for a given integer n, how many number among [1,n] are happy numbers, but counting them one by one is slow, can you help him? InputFirst line an integer t indicate there are t testcases(1≤t≤100). Then t lines follow, each line an integer n(1≤n≤106, n don‘t have leading zero). OutputOutput case number first, then the answer. Sample Input5 1 7 17 20 30 Sample OutputCase 1: 1 Case 2: 7 Case 3: 10 Case 4: 10 Case 5: 11 |
开始不敢做，怕暴力超时，后来写了暴力，但是 调代码又出问题，日
#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <algorithm>
const int N = 1010;
using namespace std;
int l=0;
int main()
{
int t,n;
scanf("%d",&t);
for(int i = 1;i<=t;i++)
{
int sum = 0,p = 0,pp = 0,ppp = 0,wz,mo;
scanf("%d",&n);
if(n<10)
{
printf("Case %d: %d\n",i,n);
continue;
}
else if(n>=10)
{
wz = 0;
int j;
for( j = 10;j<=n;j++)
{
p = j;
pp = 0,ppp = 0;
wz = 0;
while(p)
{
mo = p % 10;
wz++;
if(mo==7)
pp++;
else
ppp++;
if(ppp>=2)
break;
p /= 10;
}
if(ppp==1 && pp==wz-1)
sum++;
else if(ppp==0 && pp==wz)
sum++;
}
sum += 9;
printf("Case %d: %d\n",i,sum);
}
}
return 0;
}
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