Algorithmic Getting Started note------------Day1

The 1.C language uses%d to display the float value, does not convert the float value to an approximate int value, but instead displays the garbage value, uses%f to display the int value, and does not convert the int value to a floating-point value

2. Three-digit reversal: Enter a three-digit number, separate its hundred, 10-bit and single-digit, invert output

#include <stdio.h>int main (void) {        int A;        scanf ("%d", &a);        printf ("%d%d%d\n", a%10,a/10%10,a/100);        return 0;} Consider the case of 250, which is sufficient to output 52 or 052#include<stdio.h>int main (void) {        int A;        scanf ("%d", &a);        a=a%10*100+a/10%10*10+a/100;        printf ("%03d\n", a);        return 0;}

3. Two-digit exchange

Variable Exchange #include<stdio.h>int main (void) {        int t,a,b;        scanf ("%d%d", &a,&b);        T=a;        a=b;        b=t;        printf ("%d%d", A, b);        return 0;} The two-variable method #include<stdio.h>int main (void) {int A a        , A;        scanf ("%d%d", &a,&b);        A=a+b;        B=a-b;        A=a-b;        printf ("%d%d\n" A, b);} Different or thought, be cautious with # define swap (A, b) a^=b^=a^=b Note here that the two modifications to the same variable cannot be placed in an expression, #include <stdio.h> #define SWAP (A, b) a^=b^= A^=bint Main (void) {int A=1,b=2;int c[]={1,2};swap (A, b), swap (c[0],c[1]);p rintf ("a=%d,b=%d\n", A, b);p rintf ("c[0]=%d, C[1]=%d\n ", c[0],c[1]); return 0;} Running results under GCC: a=2,b=1c[0]=0,c[1]=1 running results in VC6.0 environment: a=2,b=1c[0]=2,c[1]=1

43 integers

#include <stdio.h>int main (void) {        int a,b,c,t;        scanf ("%d%d%d", &a,&b,&c);        if (a>b)  {            t=a;a=b;b=t;        }        if (a>c) {            t=a;a=c;c=t;        }        if (b>c) {            t=b;b=c;c=t;        }        From small to large arrange        printf ("%d%d%d\n", a,b,c);        return 0;} Note that if this is a, B, and then compare B,c, the last a,c, here it is possible that the median was modified two times #include<stdio.h>int main (void) {        int a,b,c,x,y,z;        scanf ("%d%d%d", &a,&b,&c);        X=a;if (b<x)  x=b;if (c<x)  x=c;        Z=a;if (b>z)   z=b;if (c>z)   z=c;        Y=a+b+c-x-z;        printf ("%d%d%d\n", X, Y, z);        return 0;}

Trigonometric functions in 5.C language use radians instead of angles

6. Leap Year judgment

if (year%4==0&&year%100!=0) | | (year%400==0)
return 1;
Else
return 0;

7. Four-bit complete square number with output shape such as Aabb

#include <stdio.h> #include <math.h>int main (void) {        int a,b,s;        for (a=1;a<=9;a++) for          (b=0;b<=9;b++)            {                    s=1100*a+11*b;                    if (Floor (sqrt (s) +0.5) ==sqrt (s))   //floor is the integer portion of the return x, in order to reduce the error by rounding floor (x+0.5)                        printf ("%d\n", s);            }        return 0;} #include <stdio.h>int main (void) {        int x,n,hi,ho;        for (x=1;; x + +)        {            n=x*x;            if (n<1000)  continue;            if (n>9999) break  ;            hi=n/100;            ho=n%100;            if (hi/10==hi%10&&ho/10==ho%10)   printf ("%d\n", n);        }        return 0;}

8.3n+1 problems

#include <stdio.h>int main (void) {        int n,num=0;        scanf ("%d", &n);        while (n>1)        {            if (n%2==1)      n=3*n+1; Big data, multiplication can overflow            else        n=n/2;            num++;        }        printf ("%d\n", num);        return 0;} When the output 987654321 error, it is the multiplication overflow//In order to solve this problem, there is a good way to solve, that is, when n is odd, the 3n+1 must be even, when immediately divided by 2, that is (3n+1)/2= (2n+n+1)/2=n+ (n +1)/2; #include <stdio.h>int main (void) {        int n,num=0;        scanf ("%d", &n);        while (n>1)        {            if (n%2==1)      {                n=n+ (n+1)/2;    Modify here                num++;            }            else        N=N/2;            num++;        }        printf ("%d\n", num);        return 0;}

9. The sum of factorial

#include <stdio.h>int main (void) {        int i,j,n;        int s=0;        scanf ("%d", &n);        for (i=1;i<=n;i++)        {                int fac=1;                for (j=1;j<=i;j++)                    fac*=j;                S+=FAC;        }        printf ("%d\n", s%1000000);        return 0;} This may overflow, because the factorial data is very large, so the use of each step to the N-redundancy, the result is not changed, this is a knowledge point of number theory #include<stdio.h> #include <time.h>int main (void) {        int i,j,n;        int s=0;        const int mod=1000000;        scanf ("%d", &n);        for (i=1;i<=n;i++)        {                int fac=1;                for (j=1;j<=i;j++)                    fac= (fac*j%mod);               S= (S+FAC)%mod;        }        printf ("%d\n", s);        printf ("Time used=%.2lf\n", (double) clock ()/clocks_per_sec); Learn the usage of the time header file        return 0;}

10. File operation

#define Local#include<stdio.h> #define INF 1000000000int Main (void) {#ifdef LOCAL freopen ("data.in"            , "R", stdin);        Freopen ("Data.out", "w", stdout);        #endif//LOCAL;        int x=0,n,min=inf,max=-inf,s=0;                while (scanf ("%d", &n) ==1) {if (N>max) max=n;                if (n<min) min=n;                S+=n;        x + +;        } printf ("%d%d%.3lf\n", Min,max, (double) s/x); return 0;}        File input/output operation, not using redirection #include<stdio.h> #define INF 10000000int Main (void) {file *fin,*fout;        Fin=fopen ("data.in", "RB");        Fout=fopen ("Data.out", "WB");        int x=0,n,s=0,min=inf,max=-inf;            while (FSCANF (Fin, "%d", &n) ==1) {if (N>max) max=n;            if (n<min) min=n;            S+=n;        x + +;        } fprintf (Fout, "%d%d%.3lf\n", Min,max, (double) s/x);        Fclose (Fin);        Fclose (Fout); return 0;}

Exercises

Enter an integer that does not exceed 10 9, and the number of bits to output it #include<stdio.h>int main (void) {int n;        scanf ("%d", &n);        int num=1;                while (n>9) {N=N/10;        num++; } printf ("%d\n", num);}        Narcissus number #include<stdio.h>int main (void) {int a,b,c;            for (int i=100;i<=999;i++) {a=i/100;            b=i/10%10;            c=i%10;        if (a*a*a+b*b*b+c*c*c==i) printf ("%d\n", I); } return 0;}        Han Xin Soldiers of #include<stdio.h>int main (void) {int a,b,c,i;        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        } else continue;        } if (i>100) printf ("No answer\n"); return 0;}        Inverted triangle #include<stdio.h>int Main (void) {int n;        scanf ("%d", &n); for (int i=5;i>=1;i--) {for (int j=1;j<=2*i-1;j++) printf ("#");                printf ("\ n");        for (int k=5;k>=i;k--) printf (""); } return 0;}        Fractional decimal #include<stdio.h>int main (void) {int a,b,c;        scanf ("%d%d%d", &a,&b,&c);        printf ("%.*lf\n", C, (double) A/b); return 0;}        Approximate calculation, a bit of a problem #include<stdio.h> #include <math.h>int main (void) {int t=-1;        Double a=1.0,sum=1.0;                while (Fabs (a) >=0.01) {a= (1.0)/(a+2);                A=a*t;                Sum=sum+a;        T=t*-1;        } printf ("%.9lf\n", sum); return 0;}

　　

Algorithmic Getting Started note------------Day1