题目链接 Given an array of digits which is sorted in non-decreasing order. You can write numbers using each digits[i] as many times as we want. For example, if digits = ['1','3','5'], we may write number
Given an integer n, count the total number of digit 1 appearing in all non-negative integers less than or equal to n. For example: Given n = 13, Return 6, because digit 1 occurred in the followin
Find the nth digit of the infinite integer sequence 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ... Note: n is positive and will fit within the range of a 32-bit signed integer (n < 231).
【POJ3187】【洛谷1118】Backward Digit Sums Time Limit : 2000/1000ms (Java/Other) Memory Limit : 131072/65536K (Java/Other) Problem Description FJ and his cows enjoy playing a mental game. They write do
Question: Given an integer n, count the total number of digit 1 appearing in all non-negative integers less than or equal to n. For example: Given n = 13, Return 6, because digit 1 occurred in th
E题:Digit Sum Divisible 标签:数位 d p dp dp题意:给定一个 n n n,求小于等于 n n n的数中有多少个能被自己的位数之和整除。( 1 < = n < = 1 0 14 1<=n<=10^{14} 1<=n<=1014)**题解:**数位 d p dp dp模版题, d p [ p o s ] [ s u m ] [ m o d ] dp[pos][sum][
Given an integer n, count the total number of digit 1 appearing in all non-negative integers less than or equal to n. For example: Given n = 13, Return 6, because digit 1 occurred in the following
Given an integer n, count the total number of digit 1 appearing in all non-negative integers less than or equal to n. For example: Given n = 13, Return 6, because digit 1 occurred in the followin
题目: Given an integer n, count the total number of digit 1 appearing in all non-negative integers less than or equal to n.For example:Given n = 13,Return 6, because digit 1 occurred in the following
题目:Factorial digit sum n! means n (n 1) ... 3 2 1 For example, 10! = 10 9 ... 3 2 1 = 3628800, and the sum of the digits in the number 10! is 3 + 6 + 2 + 8 + 8 + 0 + 0 = 27. Find the sum of the di
题目:Power digit sum 215 = 32768 and the sum of its digits is 3 + 2 + 7 + 6 + 8 = 26. What is the sum of the digits of the number 21000? 这个题目在于大数乘法以及减枝。 数学描述 首先有个结论,如下: 那么,N = 21000时,C(N)=30
题目链接 #include <iostream>#include <cstdio>using namespace std;int mod_exp(int a, int b, int c) //快速幂取余a^b%c{int res, t;res = 1 % c; t = a % c;while (b){if (b & 1){res = res * t % c;}t = t *