再接再厉
Number Sequence
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Time limit: 1 Seconds Memory limit: 32768K
Total Submit: 1092 Accepted Submit: 307
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A single positive integer i is given. Write a program to find the digit located in the position i in the sequence of number groups S1S2...Sk. Each group Sk consists of a sequence of positive integer numbers ranging from 1 to k, written one after another.
For example, the first 80 digits of the sequence are as follows:
11212312341234512345612345671234567812345678912345678910123456789101112345678910
Input
The first line of the input file contains a single integer t (1 <= t <= 10), the number of test cases, followed by one line for each test case. The line for a test case contains the single integer i (1 <= i <= 2147483647)
Output
There should be one output line per test case containing the digit located in the position i.
Sample Input
2
8
3
Sample Output
2
2
/*解释一下:给定一个有规律的序列,它的每一个元素是由一位数字组成的,对输入一数,代表某元素在这个序列中的位置,输出这个元素,大家注意一下,10,11等等是拆开成2(3,4...)个数字的,也就占了两个位置.*/
----------------解决方案--------------------------------------------------------
2^x mod n = 1
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Time limit: 1 Seconds Memory limit: 32768K
Total Submit: 4209 Accepted Submit: 1144
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Give a number n, find the minimum x that satisfies 2^x mod n = 1.
Input
One positive integer on each line, the value of n.
Output
If the minimum x exists, print a line with 2^x mod n = 1.
Print 2^? mod n = 1 otherwise.
You should replace x and n with specific numbers.
Sample Input
2
5
Sample Output
2^? mod 2 = 1
2^4 mod 5 = 1
/*找出满足2^x mod n==1的最小x,如果找不到,则输出时用?代替.*/
----------------解决方案--------------------------------------------------------
这样对吗？
#include <stdio.h>
#include <stdlib.h>
int main()
{
int t, j;
long s;
while(scanf("%d", &t) != EOF)
{
for(j = 0;j < t;j ++)
{
scanf("%ld", &s);
int i, k;
for(i = 1;;i ++)
{
k = (i*(i + 1))/2 - s;
if(k >= 0)
break;
}
printf("%d\n",i - k);
}
}
return 0;
}
----------------解决方案--------------------------------------------------------
第一个没这么简单吧
----------------解决方案--------------------------------------------------------
我测试了很多，都对了！
----------------解决方案--------------------------------------------------------
哦。我知道了！是不是最大的只能是10啊！
----------------解决方案--------------------------------------------------------
输入的数可以很大,但输出的数只有0 1 2 3 4 5 6 7 8 9 中的一个.
----------------解决方案--------------------------------------------------------
这样对吗？
#include <stdio.h>
#include <stdlib.h>
int main()
{
int t, j;
long s;
while(scanf("%d", &t) != EOF)
{
for(j = 0;j < t;j ++)
{
scanf("%ld", &s);
int i, k;
for(i = 1;;i ++)
{
k = (i*(i + 1))/2 - s;
if(k >= 0)
break;
}
printf("%d\n",i - k);
}
}
return 0;
}
数据用long吧
[此贴子已经被作者于2006-12-10 20:58:15编辑过]
----------------解决方案--------------------------------------------------------
第二题答案：
#include <stdio.h>
#include <math.h>
main()
{
int n;
double x;
printf("Please input a positive integer:");
scanf("%d",&n);
if(n<=0)
printf("Error input!");
else
{
if(n%2==0||n==1)
printf("2^? mod %d = 1\n",n);
else
{
for(x=2;;x++)
if((int)pow(2,x)%n==1)
{
printf("2^%.0f mod %d = 1\n",x,n);
break;
}
}
}
}
[此贴子已经被作者于2006-12-10 22:19:19编辑过]
----------------解决方案--------------------------------------------------------
12
2^? mod 12 = 1
234
2^? mod 234 = 1
21
2^6 mod 21 = 1
223
2^37 mod 223 = 1
2345
2^132 mod 2345 = 1
23457
2^1116 mod 23457 = 1
1234567
2^5670 mod 1234567 = 1
1234321
2^111100 mod 1234321 = 1
19876765
2^1987676 mod 19876765 = 1
129826321
2^32456580 mod 129826321 = 1
----------------解决方案--------------------------------------------------------