UVA - 10706 Number Sequence(贪心)
2014-09-12 19:24
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Problem B
Number Sequence
Input: standard input
Output: standard output
Time Limit: 1 second
A single positive integer iis given. Write a program to find the digit located in the position iin the sequence of number groups S1S2�Sk. Each group Skconsists
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 <=25), 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)
题目大意:给你一个升序的序列,序列由如下序列组成
1
12
123
1234
12345
123456
1234567
12345678
123456789
12345678910
1234567891011
12345678910
现在给你一个数字n,代表在第几个位置,现在请你输出这个位置的数字。
(1 <=n <=2147483647)
解析:这题主要用到贪心的思想。
先打表,开一个数组digit保存每行的数字的个数,再开一个数组sum保存到当前行所有数字的个数,因为n的值可能很大,sum要开成long long。
现在遍历sum的数组,找到第一个sum[i] >= n,计算出sum[i-1] 到 n的距离dis,再用一个cnt来计算到sum[i-1]的距离,当cnt >= dis时,求出cnt - dis 的距离,这个距离就是当前该数字的第几位,输出该位数的值。
Number Sequence
Input: standard input
Output: standard output
Time Limit: 1 second
A single positive integer iis given. Write a program to find the digit located in the position iin the sequence of number groups S1S2�Sk. Each group Skconsists
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 <=25), 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�������������������������� Output for Sample Input
2 8 3 | 2 2 |
1
12
123
1234
12345
123456
1234567
12345678
123456789
12345678910
1234567891011
12345678910
现在给你一个数字n,代表在第几个位置,现在请你输出这个位置的数字。
(1 <=n <=2147483647)
解析:这题主要用到贪心的思想。
先打表,开一个数组digit保存每行的数字的个数,再开一个数组sum保存到当前行所有数字的个数,因为n的值可能很大,sum要开成long long。
现在遍历sum的数组,找到第一个sum[i] >= n,计算出sum[i-1] 到 n的距离dis,再用一个cnt来计算到sum[i-1]的距离,当cnt >= dis时,求出cnt - dis 的距离,这个距离就是当前该数字的第几位,输出该位数的值。
#include <cstdio> #include <cstring> #include <cmath> #define ll long long using namespace std; const int N = 100000; ll sum ; ll digit ; void init() { sum[0] = 0; digit[0] = 0; for(int i = 1; i < N; i++) { digit[i] = digit[i-1] + log10(i) + 1; sum[i] = sum[i-1] + digit[i]; } } int solve(int n) { for(int i = 1; i < N; i++) { if(n <= sum[i]) { ll dis = n - sum[i-1]; int ans,cnt = 0; int j; for(j = 1; j <= i; j++) { cnt += (log10(j) + 1); if(cnt >= dis) { break; } } if(cnt == dis) { ans = j % 10; return ans; }else if(cnt > dis) { ans = j / pow(10,(cnt-dis)); ans %= 10; return ans; } } } return 0; } int main() { init(); int t,n; scanf("%d",&t); while(t--) { scanf("%d",&n); int ans = solve(n); printf("%d\n",ans); } return 0; }
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