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