2014嘉杰信息杯ACM/ICPC湖南程序设计邀请赛暨第六届湘潭市程序设计竞赛 -Happy Number

本文链接：https://blog.youkuaiyun.com/u014361775/article/details/29208359

本文介绍了一种高效方法来计算给定范围内包含所有7或仅包含一个非7数字的幸福数字的数量。通过避免逐个计数，实现算法优化。

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Happy Number


Accepted : 116 Submit : 288
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

水题  暴力就可以 但要注意两个不是7的跳出  否

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
using namespace std;
int main()
{
    int n,i,t,k;
    scanf("%d",&n);

    for(i=1;i<=n;i++)
    {
       int sum=9;
        scanf("%d",&t);
        if(t<10)
            printf("Case %d :%d\n",i,t);
        else
        {

            for(k=10;k<=t;k++)
            {
                int k1=k;
                 int p=0;10
                int t1 = 0;
                int x;
                int w=0;
                 while (k1)
                        {
                       x = k1%10;
                       k1=k1/10;
                       t1++;
                       if(x==7)
                         p++;
                       else w++;

                      if(w>=2)
                        break;
                      }



               if(p==t1||p==t1-1)
                   sum++;


          }
 printf("Case %d: %d\n",i,sum);


            }


    }

    return 0;
}