POJ 1019 数论基础题
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Number Sequence
Description
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 Source
Tehran 2002, First Iran Nationwide Internet Programming Contest
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题意：给你一个这样的数列1,12,123,1234,1234,12345,....要你求出第n位的数字是什么。思路：一开始把两位数以上的数当成只有一位数来算，然后就WA了;sum[i]表示以i结尾到前面所以数的总位数；比如sum[3]=6;(即1,12,123的位数总和);
代码：
#include
#include
#include
#include
using namespace std;
#define maxn 100007
#define inf 0x3f3f3f3f
typedef long long ll;
ll sum[40007],n;
void fun()
{
sum[0] = 0;
for(int i = 1;i<40000;i++){
if(i <10){
sum[i] = sum[i-1] + i;
}
else if(i >= 10 && i <100){
sum[i] = sum[i-1] + 9+(i-9)*2;
}
else if(i >= 100 && i <1000){
sum[i] = sum[i-1] + 9 + (99-9)*2 + (i-99)*3;
}
else if(i >= 1000 && i<10000){
sum[i] = sum[i-1] + 9 + (99-9)*2 + (999-99)*3 + (i-999)*4;
}
else if(i >= 10000 && i < 100000){
sum[i] = sum[i-1] + 9 + (99-9)*2 + (999-99)*3 + (9999-999)*4 + (i-9999)*5;
}
}
}
int main()
{
#ifdef CDZSC_June
freopen("t.txt", "r", stdin);
//freopen("E:\\w.txt", "w", stdout);
#endif
int t,p,k;
scanf("%d",&t);
fun();
while(t--)
{
scanf("%lld",&n);
for(int i = 1; ;i++)
{
if(sum[i] > n){
p = i;
break;
}
}
if(n-sum[p-1] == 0){
printf("%d\n",(p-1)%10);
}
else{
ll p1;
if(n-sum[p-1] < 10){
printf("%lld\n",n-sum[p-1]);
}
else if(n-sum[p-1] >= 10 && n-sum[p-1] <190){
p1 = n-sum[p-1] - 9;
k = (p1-1)/2;
if(p1%2==0){
printf("%d\n",(10+k)%10);
}
else{
printf("%d\n",(10+k)/10);
}
}
else if(n-sum[p-1] >= 190 && n-sum[p-1] <2890){
p1 = n-sum[p-1] - 9 - 90*2;
k = (p1-1)/3;
if(p1%3==0){
printf("%d\n",(100+k)%10);
}
else if(p1%3==1){
printf("%d\n",(100+k)/100);
}
else{
printf("%d\n",((100+k)/10)%10);
}
}
else if(n-sum[p-1] >= 2890 && n-sum[p-1] <38890){
p1 = n-sum[p-1] - 9 - 90*2 - 900*3;
k = (p1-1)/4;
if(p1%4==0){
printf("%d\n",(1000+k)%10);
}
else if(p1%4==1){
printf("%d\n",(1000+k)/1000);
}
else if(p1%4==2){
printf("%d\n",((1000+k)/100)%10);
}
else{
printf("%d\n",((1000+k)/10)%10);
}
}
else if(n-sum[p-1] >= 38890 && n-sum[p-1] <488890){
p1 = n-sum[p-1] - 9 - 90*2 - 900*3 - 9000*4;
k = (p1-1)/5;
if(p1%5==0){
printf("%d\n",(10000+k)%10);
}
else if(p1%5==1){
printf("%d\n",(10000+k)/10000);
}
else if(p1%5==2){
printf("%d\n",((10000+k)/1000)%10);
}
else if(p1%5==3){
printf("%d\n",((10000+k)/100)%10);
}
else{
printf("%d\n",((10000+k)/10)%10);
}
}
}
}
return 0;
}