C--Partial loop excerpt

1.99 Multiplication Table

#include <stdio.h>int main (int argc,const char *argv[]) {    //Outer loop--How many lines for    (int i=1; i<=9; i++) {       // Inner Loop-The specific entry for each line printed for        (int j=1; j<=i; J + +) {            printf ("%d*%d=%d", J, I, i*j);        }        printf ("\ n");    }    return 0;}

2. Alphabet Pyramid

Input: f////output://     f//    efe//   defed//  cdefedc//BCDEFEDCB//ABCDEFEDCBA
/*
Analysis:

     F

    EF  E   DEF  ED  cdef  EDC bcdef  edcbabcdef  EDCBA

*/int Main (int argc,const char *argv[]) {    char ch;    scanf ("%c", &ch);
    Outer Loop--How many lines are printed for    (int i=0; i<ch-' A ' +1; i++) {
        Print spaces-note the relationship between the number of spaces and I for        (int j=0; j<ch-' A '-i; j + +) {            printf ("");        }
        Print left half character        for (int j=i; j>=0; j--) {            printf ("%c", ch-j);        }
        Prints a half-part character        for (int j=0; j<i; J + +) {            printf ("%c", ch-1-j);        }        printf ("\ n");    }    return 0;}

    input: f//    output://    fedcba//    edcbab//    dcbabc//    cbabcd//    babcde//ABCDEF
/*
Analysis: In the same vein
*/        char ch;    scanf ("%c", &ch);    for (int i=0; i<ch-' a ' +1; i++) {        for (int j=ch-' a '-I; j>=0; j--) {            printf ("%c", ' a ' +j);        }        for (int j=0; j<i; J + +) {            printf ("%c", ' A ' +j+1);        }        printf ("\ n");    }

Input: f////output:////       a//     aba//    abcba//  abcdcba//abcdedcba//abcdefedcbaint Main (int argc,const char * Argv[]) {    char ch;    scanf ("%c", &ch);        for (int i=0, i<ch-' a ' +1; i++) {for        (int j=0; j<ch-' a '-I; j + +) {            printf ("");        }        for (int j=0; j<=i; J + +) {            printf ("%c", ' A ' +j);        }        for (int j=i; j>0; j--) {            printf ("%c", ' A ' +j-1);        }        printf ("\ n");    }    return 0;}

Input: f////output:////fedcba//edcbab//  dcbabc//   cbabcd//    babcde//     abcdefint Main (int argc,const char * Argv[]) {    char ch;    scanf ("%c", &ch);    for (int i=0, i<ch-' A ' +1; i++) {for        (int j=i; j>0; j--) {            printf ("");        }        for (int j= ch-' a '-i;j>=0; j--) {            printf ("%c", ' a ' +j);        }        for (int j=0; j<i; J + +) {            printf ("%c", ' A ' +1+j);        }        printf ("\ n");    }        return 0;}

3. (Loop inversion)

eg. given a 5-bit integer, reverse the number according to the 10 binary bit, for example a given 12345 becomes 54321

    int m;    scanf ("%d", &m);    int n=0;    while (m) {
        Remove the last digit and move the N=N*10 + m%10 to the high position        ;
        Remove the last        m=m/10;    }    printf ("%d\n", n);    return 0;

4. (Cycle through the figures)

eg. add the values of all bits of a 8-bit integer and output

    int m;    scanf ("%d", &m);    int n=0;    int sum=0;    while (m) {
        Remove each bit        of n=m%10;//printf ("%d", N) in turn;        Sum=sum+n;
        Remove the last        m=m/10;    }    printf ("%d\n", sum);

5. (Loop Solve "end number")

eg. find all "end numbers" within 1000, the so-called end number is the sum of all its factors such as: 6 = 1+2+3;

for (int i = 1; I <=, i++)    {        int sum = 0;        for (int j = 1; j < I; j + +)        {
            Find the number that matches, add            if (i% J = = 0)                sum = sum + j;        }
        Satisfies the condition        if (i = = sum)            printf ("%d\n", I);    }

6. Greatest common divisor & least common multiple

Greatest common divisor, least common multiple   //least common multiple = Two integer multiplied by greatest common divisor//division//    ①a%b remainder c//    //    ② if c=0, then B is two number of greatest common divisor/    ///③ If c≠0, then a=b,b=c, then go back to execute ①    int m,n;    int a,b,c;    scanf ("%d%d", &a,&b);    M=a;    n=b;    Go    while (b!=0) {        c=a%b;        a=b;        b=c;//remainder Assignment    }    printf ("Greatest common divisor%d\n", a);    printf ("Least common multiple%d\n", m*n/a);

eg. enter two fractions and output the sum of two fractions (numerator required)

    int a, B;    int m,n;    scanf ("%d/%d", &a,&b);    scanf ("%d/%d", &m,&n);    int x, y;    X=a*n+m*b;    Y=b*n;    Convention number    int i,j,k;    i=x;    j=y;    while (j!=0) {        k=i%j;        I=j;        j=k;    }    printf ("\ n Convention number%d\n", i);    printf ("%d/%d\n", x/i,y/i);

C--Partial loop excerpt