find the nth digit（二分查找）

题目连接：http://acm.hdu.edu.cn/showproblem.php?pid=1597

find the nth digit

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9517    Accepted Submission(s): 2757

Problem Description

假设：
S1 = 1
S2 = 12
S3 = 123
S4 = 1234
.........
S9 = 123456789
S10 = 1234567891
S11 = 12345678912
............
S18 = 123456789123456789
..................
现在我们把所有的串连接起来
S = 1121231234.......123456789123456789112345678912.........
那么你能告诉我在S串中的第N个数字是多少吗？

 

Input

输入首先是一个数字K，代表有K次询问。
接下来的K行每行有一个整数N(1 <= N < 2^31)。

 

Output

对于每个N，输出S中第N个对应的数字.

 

Sample Input

6 1 2 3 4 5 10

 

Sample Output

1 1 2 1 2 4

 

Author

8600

题解：可以用二分解，也可以用二次方程公式解，要注意使用long long

二分查找的代码：

#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
#define ll long long 
ll f(ll x)
{
    ll ans = x*(x+1)/2 ;
    return ans;
}
int main()
{
    int T;
    scanf("%d",&T);
    for(int i = 0;i < T ;i++)
    {
        ll n ;
        scanf("%lld",&n);
        if(n==1)
        {
            puts("1");
            continue;
        }
        ll l = 1;
        ll r = n;
        ll mid =( l+r )/2;
        int tm = 0;
        while(r-l>1)
        {
            if( f(mid) < n )
            {
                l = mid;
                mid = (l+r)/2;
            }
            else if( f(mid) > n)
            {
                r = mid;
                mid = (l+r)/2;
            }
            else if( f(mid)==n )
            {
                tm = mid;
                break;
            }
        }
        if(tm!=0)
        {
            if(tm%9==0) puts("9");
            else 
            printf("%d
",tm%9);
        }
        else 
        {
            ll flag = l;
            ll tt = flag*(flag+1)/2;
            int sum = n-tt;    
            if(sum%9==0) puts("9");
            else 
            printf("%d
",sum%9);
        }
    }
    return 0;
}

公式代码：

#include<cstdio>
#include<cmath>
using namespace std;
#define ll long long
#define eps 1e-8

int main()
{
    int T;
    scanf("%d",&T);
    for(int i =0 ;i < T;i++)
    {
        double n;
        scanf("%lf",&n);
        double m =(sqrt(1+8*n)-1)/2;
        ll mm =m;
        ll tm = mm*(mm+1)/2;
        ll flag = n-tm;
        if(flag == 0){if(mm%9==0)printf("9
");else   printf("%lld
",mm%9);}
        else
        {
            if(flag%9==0) printf("9
");
            else 
            printf("%lld
",flag%9);
        }
    }
    return 0 ;
}