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;
}
Answer:1366C
ompleted
on Sun, 17 Nov 2013, 15:23
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;
}
Answer:21124
Completed
on Sun, 17 Nov 2013, 16:30
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.
- 1 Jan 1900 was a Monday.
- Thirty days has September, April, June and November. All the rest have
thirty-one, Saving February alone, Which has twenty-eight, rain or shine. And
on leap years, twenty-nine. - A leap year occurs on any year evenly divisible by 4, but not on a century
unless it is divisible by 400.
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。
- 1900年1月1日是星期一。
- 30天的月份有：9月，4月，6月，11月。
- 此外的月份都是31天，当然2月除外。
- 2月在闰年有29天，其他时候有28天。
- 年份可以被4整除的时候是闰年，但是不能被400整除的世纪年（100的整数倍年）除外。
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;
}
Answer:648
Completed
on Sun, 13 Apr 2014, 02:47