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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?
题目大意:
215 = 32768 并且其各位之和为 is 3 + 2 + 7 + 6 + 8 = 26.
21000 的各位数之和是多少？
#include <stdio.h> #include <stdbool.h> void solve(void) { int a[100000] = {0}; int n, sum, i, j; n = sum = 0; a[0] = 1; for(i = 0; i < 1000; i++) { //以1000进制的方法存储 for(j = 0; j <= n; j++) { a[j] *= 2; } for(j = 0; j <= n; j++) { if(a[j] >= 10000) { a[j] %= 10000; a[j+1]++; n++; } } } for(i = 0; i <= n; i++) { sum += a[i] / 10000; a[i] %= 10000; sum += a[i] / 1000; a[i] %= 1000; sum += a[i] / 100; a[i] %= 100; sum += a[i] / 10; a[i] %= 10; sum += a[i]; } printf("%d\n",sum); } int main(void) { solve(); return 0; }
If the numbers 1 to 5 are written out in words: one, two, three, four, five, then there are 3 + 3 + 5 + 4 + 4 = 19 letters used in total.
If all the numbers from 1 to 1000 (one thousand) inclusive were written out in words, how many letters would be used?
NOTE: Do not count spaces or hyphens. For example, 342 (three hundred and forty-two) contains 23 letters and 115 (one hundred and fifteen) contains 20 letters. The use of "and" when writing out numbers is in compliance with British usage.
题目大意：
如果用英文写出数字1到5: one, two, three, four, five, 那么一共需要3 + 3 + 5 + 4 + 4 = 19个字母。
如果数字1到1000（包含1000）用英文写出，那么一共需要多少个字母？
注意: 空格和连字符不算在内。例如，342 (three hundred and forty-two)包含23个字母； 115 (one hundred and fifteen)包含20个字母。"and" 的使用与英国标准一致。
#include <stdio.h> #include <stdbool.h> int a[101] = {0,3,3,5,4,4,3,5,5,4,3,6,6,8,8,7,7,9,8,8}; void init(void) //初始化数组 { a[20] = 6; a[30] = 6; a[40] = 5; a[50] = 5; a[60] = 5; a[70] = 7; a[80] = 6; a[90] = 6; a[100] = 7; } int within100(void) //计算1~99所含字母的和 { int i, sum, t; t = sum = 0; for(i = 1; i <= 9; i++) t += a[i]; for(i = 1; i <= 19; i++) sum += a[i]; for(i = 2; i <= 9; i++) { sum += a[i*10] * 10; sum += t; } return sum; } void solve(void) { int i; int sum, t; sum = t = within100(); for(i = 1; i < 10; i++) { sum += (a[i] + 10) * 99 + (a[i] + 7) + t; } sum += 11; printf("%d\n",sum); } int main(void) { init(); solve(); return 0; }
By starting at the top of the triangle below and moving to adjacent numbers on the row below, the maximum total from top to bottom is 23.
3
7 4
2 4 6
8 5 9 3
That is, 3 + 7 + 4 + 9 = 23.
Find the maximum total from top to bottom of the triangle below:
NOTE: As there are only 16384 routes, it is possible to solve this problem by trying every route. However, Problem 67, is the same challenge with a triangle containing one-hundred rows; it cannot be solved by brute force, and requires a clever method! ;o)
题目大意：
找出从以下三角形的顶端走到底端的最大总和：
#include<stdio.h> #define N 15 int main() { char t[5]; int s[N][N]={0}; FILE *f; int i,j; f = fopen("18.txt","r"); for (i = 0; i < N; i++) { for (j = 0; j <= i; j++) { fgets(t,4,f); s[i][j] =atoi(t); } } fclose(f); for ( i = N-2; i >=0; i--) { for ( j = 0; j <= i; j++) { if (s[i+1][j] > s[i+1][j+1]) { s[i][j]+=s[i+1][j]; } else { s[i][j]+=s[i+1][j+1]; } } } printf("answer: %d\n",s[0][0]); return 0; }
Answer:1074
Completed on Thu, 1 May 2014, 16:31
You are given the following information, but you may prefer to do some research for yourself.
How many Sundays fell on the first of the month during the twentieth century (1 Jan 1901 to 31 Dec 2000)?
题目大意：
以下是一些已知信息，但是或许你需要自己做一些其他的调查。
20世纪（1901年1月1日到2000年12月31日）一共有多少个星期日落在了当月的第一天？
#include <stdio.h> #include <stdbool.h> const int a[2][12] = {{31,28,31,30,31,30,31,31,30,31,30,31}, {31,29,31,30,31,30,31,31,30,31,30,31}}; bool leapYear(int n) //判断闰年 { return (((n % 4 ==0) && (n % 100 !=0)) || (n % 400 == 0)); } bool issunday(int n) //判断某天是否是星期天 { return (n % 7 == 0 ? true : false); } void solve(void) { int num, i, j, count; count = 0; i = 1901; num = 1; while(i < 2000) { int t = (leapYear(i) ? 1 : 0); //判断闰年 for(j = 0; j < 12; j++) { num += a[t][j]; if(issunday(num)) count++; } i++; } printf("%d\n",count); } int main(void) { solve(); return 0; }
Answer:171
Completed on Mon, 18 Nov 2013, 03:38
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 digits in the number 100!
题目大意：
n! = n (n 1) ... 3 2 1
例如， 10! = 10 9 ... 3 2 1 = 3628800, 那么10!的各位之和就是3 + 6 + 2 + 8 + 8 + 0 + 0 =
27.
算出100!的各位之和。
#include <stdio.h> #include <math.h> #define N 100 int main(void){ int n=N*log(N/3),a[n],ca=0,i,j; for(i = 0; i < n; i++) a[i] = 0; a[n-1] = 1; for(i = 1; i <= N; i++){ for(j = n - 1; j >= 0; j--){ ca = i * a[j] + ca; a[j] = ca % 10; ca /= 10; } ca = 0; } ca = 0; for(i = 0; i < n; i++) ca += a[i]; printf("%d\n",ca); return 0; }
标签：style blog class code java tar
原文地址：http://www.cnblogs.com/archimedes/p/3703509.html
