Consider the infinite sequence of integers: 1, 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 2, 3, 4, 5.... The sequence is built in the following way: at first the number 1 is written out, then the numbers from 1 to 2, then the numbers from 1 to 3, then the numbers from 1 to 4 and so on. Note that the sequence contains numbers, not digits. For example number 10 first appears in the sequence in position 55 (the elements are numerated from one).
Find the number on the n-th position of the sequence.
The only line contains integer n (1 ≤ n ≤ 1014) — the position of the number to find.
Note that the given number is too large, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type.
Print the element in the n-th position of the sequence (the elements are numerated from one).
3
2
5
2
10
4
55
10
56
1
题意：一个数列 1，1,2，1,2,3,1，2,3,4,1,2,3,4,5 .....以此类推给你一个数n问第n个数是多少
题解：由题知数列的数的个数为1+2+3+4+5+6+.......所以求出第n个数是哪个子序列里的即可（第一个子序列中一个数第二个字序列里两个数第三个里三个数....）
#include<stdio.h>
#include<string.h>
#include<string>
#include<math.h>
#include<algorithm>
#define LL long long
#define PI atan(1.0)*4
#define DD double
#define MAX 5100
#define mod 10007
#define dian 1.000000011
#define INF 0x3f3f3f
using namespace std;
int main()
{
LL n,m,j,i,k,t,ans;;
while(scanf("%lld",&n)!=EOF)
{
ans=0;
for(i=1;i<2*sqrt(n)+1;i++)
{
if((i*(i+1)/2)>=n)
{
ans=i;
break;
}
}
m=ans*(ans-1)/2;
m=n-m;
printf("%lld
",m);
}
return 0;
}