C语言程序设计课后习题答案(第四版)谭浩强

scanf(\

while (score>100||score<0) {printf(\输入有误,请重输\ scanf(\ }

switch((int)(score/10)) {case 10: case 9: grade='A';break; case 8: grade='B';break; case 7: grade='C';break; case 6: grade='D';break; case 5: case 4: case 3: case 2: case 1: case 0: grade='E'; }

printf(\成绩是 %,相应的等级是%c\\n \ return 0; } 4-9

#include <> #include <> int main() {

int num,indiv,ten,hundred,thousand,ten_thousand,place; return 0; } 5-6

#include <> int main()

{double s=0,t=1; int n;

for (n=1;n<=20;n++) {

t=t*n; s=s+t; }

.=%d\\n\ printf(\ return 0; } 5-7

#include <> int main() {

int n1=100,n2=50,n3=10; double k,s1=0,s2=0,s3=0;

for (k=1;k<=n1;k++) /*计算1到100的和*/ {s1=s1+k;}

for (k=1;k<=n2;k++) /*计算1到50各数的平方和*/ {s2=s2+k*k;}

for (k=1;k<=n3;k++) /*计算1到10的各倒数和*/ {s3=s3+1/k;}

printf(\ return 0; } 5-8

#include <> int main() {

int i,j,k,n;

printf(\ for (n=100;n<1000;n++) {

i=n/100; j=n/10-i*10; k=n;

if (n==i*i*i + j*j*j + k*k*k) printf(\ }

printf(\ return 0; }

5-9-1

#define M 1000 /*定义寻找范围*/ #include <> int main()

{

int k1,k2,k3,k4,k5,k6,k7,k8,k9,k10; int i,a,n,s;

for (a=2;a<=M;a++) /* a是2-1000之间的整数，检查它是否完数 */ {n=0; /* n用来累计a的因子的个数 */

s=a; /* s用来存放尚未求出的因子之和，开始时等于a */ for (i=1;i

if (s==0) {

printf(\

if (n>1) printf(\ /* n>1表示a至少有2个因子 */ if (n>2) printf(\ /* n>2表示至少有3个因子，故应再输出一个因子 */

if (n>3) printf(\ /* n>3表示至少有4个因子，故应再输出一个因子 */

if (n>4) printf(\ /* 以下类似 */ if (n>5) printf(\ if (n>6) printf(\

if (n>7) printf(\ if (n>8) printf(\ if (n>9) printf(\ printf(\ } }

return 0; }

5-9-2

#include <> int main() {int m,s,i;

for (m=2;m<1000;m++) {s=0;

for (i=1;i

if ((m%i)==0) s=s+i; if(s==m)

{printf(\ for (i=1;i

#include <> int main() {

int i,n=20;

double a=2,b=1,s=0,t; for (i=1;i<=n;i++) {

s=s+a/b; t=a, a=a+b, b=t; }

printf(\ return 0; }

5-11

#include <> int main() {

double sn=100,hn=sn/2; int n;

for (n=2;n<=10;n++) {

sn=sn+2*hn; /*第n次落地时共经过的米数*/ hn=hn/2; /*第n次反跳高度*/ }

printf(\第10次落地时共经过%f米\\n\ printf(\第10次反弹%f米\\n\ return 0; } 5-12

#include <> int main() {

int day,x1,x2; day=9; x2=1;

while(day>0)

{x1=(x2+1)*2; /*第1天的桃子数是第2天桃子数加1后的2倍.*/ x2=x1; day--; }

printf(\ return 0; } 5-13

#include <> #include <> int main() {

float a,x0,x1;

printf(\ scanf(\ x0=a/2;