偶看新闻公园道一号郑州:c语言答案
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Ch31.#include "stdio.h"main(){int n,i;double r,p=1;scanf("%d%lf",&n,&r);for(i=1;i<=n;i++)p*=(1+r);p-=1;p*=100;printf("increase:%lf%%",p);}2.(1)#include "stdio.h"main(){int="" n;double="" r,p="1000;scanf("%d%lf",&n,&r);p*=(1+n*r);p-=1000;printf("interest:%lf",p);}(2)(3)#include" "stdio.h"main(){int="" n;double="" r,p="1000;scanf("%d%lf",&n,&r);p*=(1+n*r);scanf("%d%lf",&n,&r);p*=(1+n*r);p-=1000;printf("interest:%lf",p);}(4)#include" "stdio.h"main(){="" int="" n,i;double="" r,p=""><=n;i++)p*=(1+r);p-=1000;printf("interest:%lf",p);}(5)#include "stdio.h"main(){int="" n,i;double="" r,p=""><=n*4;i++)p*=(1+r );p-="1000;printf("interest:%lf",p);}3.#include" "stdio.h"#include="" "math.h"main(){="" long="" d="300000;int" p="6000;double" r="0.01;double" x,y,m;x="(double)p/(p-d*r);y=1+r;m=log(x)/log(y);printf("month:%.1lf",m);}6.#include" "stdio.h"main(){="" char="" c1='C' ,c2='h' ,c3='i' ,c4='n' ,c5='a' ;c1+="4;c2+=4;c3+=4;c4+=4;c5+=4;putchar(c1);putchar(c2);putchar(c3);putchar(c4);putchar(c5);putchar('\n');printf("%c%c%c%c%c\n",c1,c2,c3,c4,c5);}7.#include" "stdio.h"main(){="" double="" r,h;double="" circle,area,suarea,vol,cirvol;scanf("%lf%lf",&r,&h);circle="2*3.14*r;area=3.14*r*r;suarea=area*4;vol=suarea/3;cirvol=area*h;printf("the" circle="" is:%.2lf\n",circle);printf("the="" area="" is:%.2lf\n",area);printf("the="" superficial="" area="" is:%.2lf\n",suarea);printf("the="" volume="" is:%.2lf\n",vol);printf("the="" circular="" cylinder="" volume="" is:%.2lf\n",cirvol);}8.#include="" "stdio.h"main(){="" char="" c1,c2;c1="getchar();c2=getchar();putchar(c1);putchar(c2);printf("\n");printf("%c" %c",c1,c2);printf("\n");}ch44.#include="" "stdio.h"main(){="" int="" a,b,c,max;scanf("%d%d%d",&a,&b,&c);if(a="">b)max=a;else max=b;if(max=1000){ printf("the date error,please input again:"); scanf("%d",&a);}sq=sqrt(a);printf("the square root is:%d\n",sq);}6.#include "stdio.h"#include "math.h"main(){ int x,y;scanf("%d",&x);if(x<1) y="x;else" if(x="">=10) y=3*x-11; else y=2*x-1;printf("x=%d,y=%d\n",x,y);}8.#include "stdio.h"#include "math.h"main(){ int score,g;scanf("%d",&score);g=score/10;switch(g){case 10:case 9:printf("you grade is A");break;case 8:printf("you grade is B");break;case 7:printf("you grade is C");break;case 6:printf("you grade is D");break;case 5:case 4:case 3:case 2:case 1:case 0:printf("you grade is E");break;default:printf("you input is error!");break;}}9.解法一：#include "stdio.h"main(){ long date;int a,b,c,d,e;scanf("%ld",&date);a=date/10000;b=date%10000/1000;c=date%1000/100;d=date%100/10;e=date%10;if(a!=0) printf("the date include 5 numbers,%ld %d%d%d%d%d",date,e,d,c,b,a);else if(b!=0) printf("the date include 4 numbers,%ld %d%d%d%d",date,e,d,c,b); else if(c!=0) printf("the date include 3 numbers,%ld %d%d%d",date,e,d,c); else if(d!=0) printf("the date include 2 numbers,%ld %d%d",date,e,d); else printf("the date include 1 numbers,%ld %d",date,e);}解法二：#include "stdio.h"main(){ long date,ge;int num=0;scanf("%ld",&date);printf("%ld\n",date);while(date!=0){ge=date%10;date/=10;num++;printf("%d",ge);}printf("\nthe date include %d numbers.\n",num);}解法三：#include "stdio.h"main(){ long date,ge,dateni=0,dat;int num=0;scanf("%ld",&date);dat=date;while(date!=0){ge=date%10;date/=10;num++;dateni=dateni*10+ge;}printf("the date include %d numbers.\n",num);printf("yuanxu:%ld\n",dat);printf("nixu:%ld\n",dateni);}解法四：#include main() { long num; int indiv,ten,hundred,thousand,ten_thousand,place; scanf("%ld",&num); if (num>9999) place=5; else if(num>999) place=4; else if(num>99) place=3; else if(num>9) place=2; else place=1; printf("place =%d\n", place); ten_thousand=num/10000; thousand=num/1000%10; hundred=num/100%10; ten=num%100/10; indiv=num%10; switch(place) { case 5: printf("%d,%d,%d,%d,%d",ten_thousand,thousand,hundred,ten,indiv); printf("\n反序数字为；"); printf("%d%d%d%d%d\n",indiv,ten,hundred,thousand,ten_thousand); break; case 4: printf("%d,%d,%d,%d",thousand,hundred,ten,indiv); printf("\n反序数字为："); printf("%d%d%d%d\n",indiv,ten,hundred,thousand); break; case 3: printf("%d,%d,%d",hundred,ten,indiv); printf("\n反序数字为："); printf("%d%d%d\n",indiv,ten,hundred); break; case 2: printf("%d,%d",ten,indiv); printf("\n反序数字为："); printf("%d%d\n",indiv,ten); break; case 1: printf("%d",indiv); printf("\n反序数字为："); printf("%d\n",indiv); break; } } 10.解法一：#include "stdio.h"main(){ long lirun;double jiangjin;scanf("%ld",&lirun);if(lirun<100000) jiangjin="lirun*0.1;else"><200000) jiangjin="100000*0.1+(lirun-100000)*0.075;" else=""><400000) jiangjin="100000*0.1+100000*0.075+(lirun-200000)*0.05;" else=""><600000) jiangjin="100000*0.1+100000*0.075+200000*0.05+(lirun-400000)*0.03;" else=""><1000000) jiangjin="100000*0.1+100000*0.075+200000*0.05+200000*0.03+(lirun-600000)*0.015;" else="" jiangjin="100000*0.1+100000*0.075+200000*0.05+200000*0.03+400000*0.015+(lirun-1000000)*0.01;" printf("\nthe="" total="" jiangjin="" is:="" %lf\n",jiangjin);}解法二：#include="" "stdio.h"main(){="" long="" lirun;int="" li;double="" jiangjin;scanf("%ld",&lirun);li="lirun/100000;switch(li){case" 0:jiangjin="lirun*0.1;break;case" 1:jiangjin="100000*0.1+(lirun-100000)*0.075;break;case" 2:case="" 3:jiangjin="100000*0.1+100000*0.075+(lirun-200000)*0.05;" break;case="" 4:case="" 5:jiangjin="100000*0.1+100000*0.075+200000*0.05+(lirun-400000)*0.03;break;" case="" 6:case="" 7:case="" 8:case="" 9:jiangjin="100000*0.1+100000*0.075+200000*0.05+200000*0.03+(lirun-600000)*0.015;break;" case="" 10:default:jiangjin="100000*0.1+100000*0.075+200000*0.05+200000*0.03+400000*0.015+(lirun-1000000)*0.01;break;" }="" printf("\nthe="" total="" jiangjin="" is:="" %lf\n",jiangjin);}解法三：#include=""> main() { long i; float bonus,bon1,bon2,bon4,bon6,bon10; bon1=100000*0.1; /*利润为10万元时的奖金*/ bon2=bon1+100000*0.075; /*利润为20万元时的奖金*/ bon4=bon2+200000*0.05; /*利润为40万元时的奖金*/ bon6=bon4+200000*0.03; /*利润为60万元时的奖金*/ bon10=bon6+400000*0.015; /*利润为100万元时的奖金*/ printf("请输入利润i："); scanf("%ld",&i); if(i<=100000) bonus="i*0.1;" 利润在10万元以内按0.1提成奖金*/="" else=""><=200000) bonus="bon1+(i-100000)*0.075;" 利润在10万至20万元时的奖金*/="" else=""><=400000) bonus="bon2+(i-200000)*0.05;" 利润在20万至40万元时的奖金*/="" else=""><=600000) bonus="bon4+(i-400000)*0.03;" 利润在40万至60万元时的奖金*/="" else=""><=1000000) bonus="bon6+(i-600000)*0.015;" 利润在60万至100万元时的奖金*/="" else="" bonus="bon10+(i-1000000)*0.01;" 利润在100万元以上时的奖金*/="" printf(”奖金是%10.2f\n",bonus);="" }="" 解法四：#include=""> main() { long i; float bonus, bon1, bon2, bon4, bon6, bon10; int c; bon1=100000*0.1; bon2=bon1+100000*0.075; bon4=bon2+200000*0.05; bon6=bon4+200000*0.03; bon10=bon6+400000*0.015; printf("请输入利润i："); scanf("%ld",&i); c=i/100000; if(c>10) c=10; switch(c) { case 0: bonus=1*0.1;break; case 1: bonus=bon1+(i-100000)*0.075;break; case 2 : case 3: bonus=bon2+(i-200000)*0.05; break; case 4: case 5: bonus=bon4+(i-400000)*0.03;break; case 6: case 7: case 8: case 9: bonus=bon6+(i-600000)*0.015;break; case 10: bonus=bon10+(i-1000000)*0.01; } printf("奖金是%10.2f",bonus); } 11.#include main() { int t,a,b,c,d; printf("please input 4 int numbers："); scanf("%d%d%d%d",&a,&b,&c,&d); if(a>b) {t=a; a=b; b=t;} if(a>c) {t=a; a=c; c=t;} if(a>d) {t=a; a=d; d=t;} if(b>c) {t=b; b=c; c=t;} if(b>d) {t=b; b=d; d=t;} if(c>d) {t=c; c=d; d=t;} printf("the result：\n"); printf("%d %d %d %d\n",a,b,c,d); }12.#include "stdio.h"#include "math.h"main(){ double x,y,l,h,a,b;scanf("%lf%lf",&x,&y);a=fabs(x)-2;b=fabs(y)-2;l=sqrt(a*a+b*b);if(l<=1) h="(1-l)*10;else" h="0;printf("h=%lf\n",h);}Ch53.#" include="" "stdio.h"main()="" {long="" m,n,i,j,s,t;="" scanf("%ld,%ld",&m,&n);="" if(m="">n){t=m;m=n;n=t;}for(i=m;i>0;i--) if(m%i==0&&n%i==0) {s=i;break;}for(j=n;;j++) if(j%m==0&&j%n==0){t=j;break;}printf("gongyue=%ld,gongbei=%ld\n",s,t); }4.#include"stdio.h" main() {char c;int i=0,j=0,k=0,l=0; while((c=getchar())!='\n') {if(c>=65&&c<=90||c>=97&&c<=122) i++;="" else="" if(c="">=48&&c<=57) j++;="" else="" if(c="=32)" k++;="" else="" l++;}="" printf("i="%d,j=%d,k=%d,l=%d\n",i,j,k,l);" }5.#include"stdio.h"="" #include"math.h"="" main()="" {int="" n,a,sum="0,i=1,s;" scanf("%d%d",&n,&a);="" s=""><=n) {sum="sum+s;s=s+a*pow(10,i);" i++;}="" printf("sum="%d\n",sum);" }6.解法一：#include"stdio.h"="" main()="" {int="" n,i="1;long" sum="0,s=1;" scanf("%d",&n);=""><=n) {s="s*i;sum=sum+s;i++;}" printf("sum="%ld\n",sum);" }解法二：#include"stdio.h"="" main()="" {int="" n,i,j;long="" sum="0,s;" scanf("%d",&n);="" for(i=""><><=i;j++)s*=j;sum+=s;}printf("sum=%ld\n",sum); }7.#include"stdio.h"="" main()="" {int="" k;long="" s1="0,s2=0;double" s3="0,sum;" for(k=""><=100;k++) s1="s1+k;" for(k=""><=50;k++) s2="s2+k*k;" for(k=""><=10;k++) s3="s3+(double)1/k;" sum="s1+s2+s3;" printf("sum="%lf\n",sum);" }8.解法一#include"stdio.h"="" main()="" {int="" x,a,b,c;="" for(x=""><1000;x++){a=x 00;b="x%100/10;c=x%10;if(x==a*a*a+b*b*b+c*c*c)" printf("%d="" ",x);}}解法二：#include"stdio.h"="" #include"math.h"="" main()="" {int="" x,a,b,c;="" for(x=""><1000;x++){a=x 00;b="x%100/10;c=x%10;if(x==pow(a,3)+" pow(b,3)+="" pow(c,3))="" printf("%d="" ",x);}}9.#include"stdio.h"="" main()="" {int="" m,i,j,s;="" for(m=""><1000;m++) {s="0;" for(i="">=65&&str[i][j]<=90) k++;="" else="" if(str[i][j]="">=97&&str[i][j]<=122) x++;="" else="" if(str[i][j]="">=48&&str[i][j]<=57) y++;="" else="" if(str[i][j]="=32)" m++;="" else="" n++;}="" printf("daxie="" xiaoxie="" shuzi="" kongge="" qita\n");="" printf("%5d="" %7d="" %5d="" %6d="" %4d\n",k,x,y,m,n);="" }11.解法一：#include="" "stdio.h"="" main()="" {int="" i,j,k;char="" a[5][5];="" for(i=""><5;i++) {for(j=""><5;j++) {a[i][j]='*' ;printf("%c",a[i][j]);}="" printf("\n");="" for(k=""><=i+1;k++) printf("\40");}="" printf("\n");="" }解法二：#include="" "stdio.h"="" main()="" {int="" i,j,k;for(i=""><=5;i++) {="" for(k=""><=i-1;k++) printf("="" ");="" for(j=""><=5;j++) printf("*");="" printf("\n");="" }="" }12.#include="" "stdio.h"="" main()="" {="" int="" i;char="" str1[100],str2[100];="" gets(str1);="" for(i="0;str1[i]!='\0';i++)" if(str1[i]="">=65&&str1[i]<=90) str2[i]="155-str1[i];" else="" if(str1[i]="">=97&&str1[i]<=122) str2[i]="219-str1[i];" else="" str2[i]="str1[i];" str2[i]='\0' ;printf("%s\n%s\n",str1,str2);="" }13.#include="" "stdio.h"="" main()="" {="" int="" i,j;char="" str1[100],str2[100],str3[201];="" gets(str1);="" gets(str2);="" for(i="0;str1[i]!='\0';i++)" str3[i]="str1[i];" for(j="0;str2[j]!='\0';j++)" str3[i+j]="str2[j];" str3[i+j]='\0' ;printf("%s\n%s\n%s\n",str1,str2,str3);="" }14.#include=""> #include main() { int i,resu; char s1[100],s2[100]; printf("\n input string1："); gets(s1); printf("\n Input string2："); gets(s2); i=0; while(s1[i]==s2[i]&&s1[i]!='\0') i++; if(s1[i]=='\0'&&s2[i]=='0') resu=0; else resu=s1[i]-s2[i]; printf("\n result：%d\n",resu); }15.#include "stdio.h" main() { char s1[80],s2[80]; int i; printf("Input s2："); scanf("%s",s2); for(i=0;i=1;i--) {if(m%i==0&&n%i==0) {t=i; break;}} return(t); } int minbeishu(int m,int n) {int i,t,k; if(m>=n) k=m; else k=n;i=k; while(1){if(i%m==0&&i%n==0) {t=i;break;}i++;} return t; } main() {int a,b,max,min; printf("enter two number is: "); scanf("%d,%d",&a,&b); max=maxyueshu(a,b); min=minbeishu(a,b); printf("max=%d,min=%d\n",max,min); }解法二：#include "stdio.h" int maxyueshu(int m,int n) { int t,r; if(m>n) {t=m;m=n;n=t;}r=n%m;while(r!=0){n=m;m=r;r=n%m;}return m; } int minbeishu(int m,int n) {int t,k; k=maxyueshu(m,n);t=m*n/k;return t; } main() {int a,b,max,min; printf("enter two number is: "); scanf("%d,%d",&a,&b); max=maxyueshu(a,b); min=minbeishu(a,b); printf("max=%d,min=%d\n",max,min); }2.#include"stdio.h"#include"math.h"float shigen(float a,float b,float q) {float x1,x2; x1=(-b+sqrt(q))/(2*a); x2=(-b-sqrt(q))/(2*a); printf("two shigen is x1=%.3f and x2=%.3f\n",x1,x2); } float denggen(float a,float b) {float x; x=-b/(2*a); printf("denggen is x=%.3f\n",x); } float xugen(float a,float b,float q) {float x,y; x=-b/(2*a); y=sqrt(-q)/(2*a); printf("two xugen is x1=%.3f+%.3fi and x2=%.3f-%.3fi\n",x,y,x,y); } main() {float a,b,c,q; printf("input a b c is: "); scanf("%f,%f,%f",&a,&b,&c); printf("\n"); q=b*b-4*a*c; if(q>0) shigen(a,b,q); else if(q==0) denggen(a,b); else xugen(a,b,q); }3.#include"stdio.h"int sushu(int m) {int i,t; for(i=2;i
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