UVA - 10706 Number Sequence 子序列

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#include<cstdio>#include<cmath>#include<algorithm>using namespace std;const int maxn = 32761;long long S[maxn];//存放的是S1，S2，到SK的和，S[5]表示了S1到S4的和,当数字变化到K的时候，一共有多少个字数了int border[9] = {0,1,10,100,1000,10000,100000,1000000,10000000};int num[9] = {0,9,90,900,9000,90000,900000,9000000,90000000};//num[1]存放1-9的位数，num[2]存放1-99的位数，以此类推，刚开始初始话成这样，是为了后面的计算int getdigit(int n) {int count = 0;while(n > 0) {count++;n = n / 10;}return count;}void init() {for(int i = 2; i < 9; i++)//位数的变化，num[2]存放的是1-99的位数，因为十位数要占两位，所以要用90个十位数＊2，再加上前面的9个个位数，就是1-99的位数了，以此类推，num里面的下标表示的位数num[i] = num[i-1] + num[i] * i;S[0] = 0;int digit;for(int i = 1; i < maxn; i++) {digit = getdigit(i);S[i] = (i - border[digit] + 1 )* digit + num[digit-1];S[i] =  S[i] + S[i-1];}}int main() {int n;int N;scanf("%d",&n);init();while(n--) {scanf("%d",&N);long long *p = lower_bound(S,S+maxn,N);if(*p == N) printf("%d\n",(p-S) % 10);else {int len = N - *(p-1);bool flag = false;int i;for(i = 0; i < 9; i++) {if(num[i] == len) {flag = true;break;}else if(num[i] > len)break;}if(flag){printf("9\n");continue;}int last = len - num[i-1];int k;int count = i;for(k = border[i]; count < last; count += i,k++);;int time = count - last;while(time > 0) {k = k / 10;time--;}printf("%d\n",k % 10);}}return 0;}
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