"Introduction to the algorithm contest" on the computer Practice--Chapter II _c

Exercise 2-1 digits (digit)

Enter a positive integer that does not exceed the 10^9, outputting its number of digits. For example, 12735 digits are 5. Do not use any mathematical functions, only arithmetic and circular statements are implemented.

#include <stdio.h>

int main ()
{
    int x,num=0;
    scanf ("%d", &x);
    while (x)
    {
    	x/=10;
    	num++;
    }
    printf ("%d\n", num);
	
	return 0;

Exercise 2-2 Narcissus number (daffodil)

Output the number of daffodils in the 100~999. If the 3-digit ABC satisfies abc=a^3+b^3+c^3, it is called Narcissus number. such as 153=1^3+5^3+3^3, so 153 is the number of daffodils.

#include <stdio.h>

int main ()
{
    int i,a,b,c;
    For (i=100 i<=999; i++)
    {
    	a = i%10;     /* Single-digit
    	/b = i/10%10;  * * 10 bits
    	/c = i/100;    /* Hundred */
    	if (i = = a*a*a + b*b*b + c*c*c)
    	    printf ("%d\n", I);
    }
	
	return 0;

Results: The number of three daffodils are: 153,370,371,407

Exercise 2-3 Han Soldiers of (hanxin)

It is rumored that Han Xin talent, never directly counted the number of their own troops, as long as the soldiers to three people a row, five people a row, seven people in a row to transform formation, and he only glanced at the team's Shang to know the total number. Enter 3 nonnegative integer a,b,c, representing the number of Shang per formation (a<3,b<5,c<7), the minimum number of outputs (or no solution reported). The total number of known is not less than 10, not more than 100.

Sample input: 2 1 6

Sample output: 41

Sample input: 2 1 3

Sample output: No answer

#include <stdio.h>

int main ()
{
    int i,a,b,c;
    scanf ("%d%d%d", &a,&b,&c);
    For (i=10 i<=100; i++)
    {
    	if (i%3==a && i%5==b && i%7==c) 
    	{
    		printf ("%d\n", I);
    		break;
    	}
    }
    if (i = =)
        printf ("No answer\n");
	
	return 0;

Exercise 2-4 Inverted triangle (triangle)

Enter a positive integer n<=20, outputting an n-tier inverted triangle. For example, the output of n=5 is as follows:

#########
 #######
  #####
   ###
    #

#include <stdio.h>

int main ()
{
    int n,i,j;
    scanf ("%d", &n);
    for (i=n;i>=1;i--)
    {for
    	(j=1;j<=n-i;j++)
    	    printf ("");
		for (j=1;j<=2*i-1;j++)
    	    printf ("#");
    	printf ("\ n");
    }
	
	return 0;

Exercise 2-5 Statistics (STAT)

Enter a positive integer n, then read n positive integer a1,a2,...,an, and finally read a positive integer m. The number of statistics A1,a2,...,an is less than M. Hint: If redirects and fopen can be used, which is more convenient.

This way ... Welcome advice.

Exercise 2-6 Harmonic series (harmony)

Enter positive integer n, output H (n) = 1 + 1/2 + 1/3 +...+ 1/n value, retain 3 decimal places. For example N=3, the answer is 1.833.

#include <stdio.h>

int main ()
{
    int n,i;
    Double H = 0;
    scanf ("%d", &n);
    for (I=1; i<=n; i++)
    	H + = 1.0/i;
    printf ("%.3lf\n", H);
	
	return 0;

Exercise 2-7 Approximate calculation (approximation)

Compute PI/4 = 1-1/3 + 1/5-1/7 + ... until the last item is less than 10^ (-6).

#include <stdio.h>

int main ()
{
    int i=1,flag=1;
    Double sum=0,item=1.0;
    
    while (item>=1.0/1000000)
	{
		sum = Flag*item;
		Flag *=-1;
		i + 2;
		item = 1.0/i;
    }
    
    printf ("PI =%lf\n", sum*4); 
	
	return 0;

Exercise 2-8 the and (subsequence) of the subsequence

Enter two positive integer n<m<10^6, output 1/n^2 + 1/(n+1) ^2 +...+ 1/m^2, and retain 5 decimal places. For example, when n=2,m=4 the answer is 0.42361;n=65536,m=655360, the answer is 0.00001. Note: There is a trap.

#include <stdio.h>

int main ()
{
    int i,n,m;
    Double sum=0;
    scanf ("%d%d", &n,&m);
    
    for (i=n;i<=m;i++)
    	sum + = 1.0/i/i;
    
    printf ("%.5lf\n", sum); 
	
	return 0;

The trap: When N or M is larger, I * I will overflow, cannot use 1.0/(I * i), should use 1.0/i/I. Or use a long long data type instead.

Exercise 2-9 Fractional Decimal (decimal)

Enter a positive integer a,b,c, output A/b decimal form, accurate to the decimal point C bit. A,b<=10^6,c<=100. For example, a=1,b=6,c=4 should output 0.1667.

#include <stdio.h>

int main ()
{
    int a,b,c;
    scanf ("%d%d%d", &a,&b,&c);
    printf ("%.*lf\n", c,1.0*a/b); /* New knowledge: In the format controller, * can be replaced by the variable behind. * * return
	
	0;

The problem with this approach is that when the C is large, the fractional part after the more than 10 decimal digits shows only zero, and I don't know why.

The following is a bitwise decimal method, which is completely fine. (This code is not what I wrote, the comments I added, the source of code to see the end of the blog.) ）

#include <stdio.h>
 
int main (void) 
{ 
    int a,b,c,mod,re,i,m,x,y;
    while (scanf ("%d%d%d", &a,&b,&c) ==3)
    {
        printf ("%d", A/b);   /* Output Integer part * 
        /mod=a%b;
        if (c>0)
        {
            printf (".");
            for (i=1;i<c;i++)/  * Bitwise calculation OUTPUT c-1 decimal * *
                m=mod*10;
                re=m/b;
                printf ("%d", re);
                mod=m%b;
            }
            m=mod*10;
            x=m/b;
            mod=m%b;    /* More than one, rounded to the last one we need * * *
            m=mod*10;
            y=m/b;
            if (y>=5)
            x + +;
            printf ("%d\n", x);  /* Output Last decimal *
    /}} return
    0; 
}

Exercise 2-10 Arrangement (permutation)

With 1,2,3,..., 9 consists of 3 three-digit abc,def and GHI, each number is used exactly once, requiring Abc:def:ghi = 1:2: 3. Outputs all the solutions. Tip: Don't use your brains.

#include <stdio.h> int arr[10] = {0};/* array arr to record whether 1~9 each number appears, corresponds to subscript one by one, appears as 1, otherwise 0*/int main () {int a,b,c,d,e,f,g,h,i
	, Abc,def,ghi,t,sum; for (a=1;a<=3;a++)/*a Max is 3*/for (b=1;b<=9;b++) for (c=1;c<=9;c++) for (d=2;d<=6;d++)/*d min 2, max for 6* /for (e=1;e<=9;e++) for (f=1;f<=9;f++) for (g=3;g<=9;g++)/*g min 3*/for (h=1;h<=9;h
										    
						+ +) for (i=1;i<=9;i++) {for (t=1;t<=9;t++) arr[t] = 0;
						Arr[a] = 1;arr[b] = 1;arr[c] = 1;
					    ARR[D] = 1;arr[e] = 1;arr[f] = 1;
										
						ARR[G] = 1;arr[h] = 1;arr[i] = 1;
	                    sum = 0;
	                    for (t=1;t<=9;t++)/* If the array is cumulative and is 9, which means that all 1~9 numbers appear * * * sum+=arr[t];
	                        if (sum==9) {abc = A*100+B*10+C;
	                        def = d*100+e*10+f;
			                Ghi = G*100+h*10+i; if (abc*2==def && Abc*3==ghi) printf ("%d:%d:%d = 1:2:3\n", Abc,def,ghi);
} return 0; }

Results:

192:384:576 = 1:2:3

219:438:657 = 1:2:3

273:546:819 = 1:2:3

327:654:981 = 1:2:3

There is another online method, a certain degree of reverse thinking, exquisite, running faster than my multi-layer for loop nesting more quickly, my program cycle number of 3*9*9*5*9*9*7*9*9 = 55,801,305 times, and the following procedures only 333-100 = 233 times.

#include <stdio.h>

int main (void)
{
	int x, y, z, a[10] = {0};
	for (x = n < 333 x + +)
	{
		y = 2*x;
		z = 3*x;
		Number of a[appearing] = 1
		a[x/100] = a[x/10%10] = a[x%10] = 1;
		A[Y/100] = a[y/10%10] = a[y%10] = 1;
		A[Z/100] = a[z/10%10] = a[z%10] = 1;
		int I, s = 0;
		for (i = 1; i < i++)
			s + = a[i];
		if (s = = 9)
			printf ("%d\t%d\t%d\n", X, Y, z);
		for (i = 1; i < i++)	//re-assign to 0
			a[i] = 0;
	}
	return 0;
}

If there are errors, please note that the discussion is welcome.

Note: This article is a partial reference to this blog: http://blog.csdn.net/litiouslove/article/details/7891700