#include <stdio.h>intMain () {DoubleJc=1.0; inti; for(i=1; i<= -; i++) {JC*=i; } printf ("1!+2!+...+20!=%LF", JC);} #include<stdio.h>intIsPrime (intN) { inti,result=1; for(i=2; i<n/2; i++) { if(n%i==0) {result=0; Break; } } returnresult;}intMain () {intN; scanf ("%d", &N); if(IsPrime (n)) {printf ("0", N); } Else{printf ("1", N); } return 0;} #include<stdio.h>#include<math.h>intIsPrime (intN) { inti; for(i=2; I<=sqrt (n); i++) if(n%i==0)return 0; return 1;}intMain () {inti,j,s=0; for(i=2; i<= -; i++) if(IsPrime (i)) printf ("%4d", i); return 0;} #include<stdio.h>#include<math.h>intMain () {printf ("Please enter two number: \ n"); intb; scanf ("%d%d",&a,&b); intn=A; if(n>b) N=b; for(inti=n;i>=1; i--){ if(a%i==0&&b%i==0) {printf ("Greatest common divisor:%d \ n", i); Break; } }} #include<stdio.h>intMain () {intN,m,i,s,a; printf ("enter any two positive integers:"); scanf ("%d,%d",&n,&m); for(a=m;a<=m*n;a++) if(a%m==0&&a%n==0) {printf ("%d\n", a); return 0; }} #include<stdio.h>#include<math.h>floatMax; floatMin; floatAveragefloatB[],intN) {inti; floatTotal ; Max= b[0]; Min= b[0]; Total= b[0]; for(i=1; i<n;i++) { if(B[i] >Max) {Max=B[i]; } if(B[i] <Min) {Min=B[i]; } Total= Total +B[i]; } returntotal/N; } intMain () {floatarray[Ten]; inti; floatAve; printf ("Please input 10 people's results, the middle with a space separate: \ n"); for(i=0;i<Ten; i++) {scanf ("%f",&Array[i]); } Ave= Average (Array,Ten); printf ("Highest score is:%6.2f\n", Max); printf ("the lowest score is:%6.2f\n", Min); printf ("The average score is:%6.2f\n", Ave); return 0; }
Experience: This experiment I think the most difficult is the last problem, the original solution to this problem requires three functions, but this time only used one. Just contact with this problem I feel at a loss, do not know how to start, after listening to the teacher in the course of the content and check the information based on the completion. This shows that we are still not deep enough to learn that the real ability can only rely on their own, light on the lesson is far from enough.
160809207 Liu Chaoxuan