1、a+b1.9题 int add(int x,int y); c=add(a,b);int add(int x,int y)int z; z=x+y; return(z);2.3题 char c1=a,c2=b,c3=c,c4=101,c5=116c1c2c3ntbc4tc52.4题C+I say: ttHe says:C+ is very interesting! 2.7题int i,j,m,n; i=8; j=10; m=+i+j+; n=(+i)+(+j)+m;ijmnint main ( )float h,r,l,s,sq,vq,vz; const float pi=3.1415926;
2、please enter r,h:rh; l=2*pi*r; s=r*r*pi; sq=4*pi*r*r; vq=3.0/4.0*pi*r*r*r; vz=pi*r*r*h;setiosflags(ios:fixed)right) setprecision(2);l= setw(10)ls= ssq=sqvq=vqvz=vzi2; c1=i1; c2=i2;按字符输出结果为: , 3.8题 int a=3,b=4,c=5,x,y;(a+bc & b=c)(a|b+c & b-c)b) & !c|1)(x=a) & (y=b) & 0)(a+b)+c-1 & b+c/2)3.9题include
3、int a,b,c;please enter three integer numbers: if(ab) if(bc)max= else else if (ab)?a: /* 将a和b中的大者存入temp中 */ max=(tempc)?temp: /* 将a和b中的大者与c比较,最大者存入max */max3.10题 int x,y;enter x:x;1) y=x;x=x, y=x=y; else if (x100|score0) coutdata error,enter data again. switch(int(score/10) case 10: case 9: grade=Abr
4、eak; case 8:B case 7: case 6:D default:grade=Escore is score, grade is grade9999) place=5; else if (num999) place=4;99) place=3;9) place=2; else place=1;place=place /计算各位数字 ten_thousand=num/10000; thousand=(int)(num-ten_thousand*10000)/1000; hundred=(int)(num-ten_thousand*10000-thousand*1000)/100; t
5、en=(int)(num-ten_thousand*10000-thousand*1000-hundred*100)/10; indiv=(int)(num-ten_thousand*10000-thousand*1000-hundred*100-ten*10);original order: switch(place) case 5:coutten_thousandthousandhundredtenindivendl;reverse order: break; case 4: case 3: case 2: case 1: 3.13题 long i; /i为利润 float bonus,b
6、on1,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+100000*0.05; /利润为40万元时的奖金 bon6=bon4+100000*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enter i:i; if (i=100000) bonus=i*0.1; /利润在10万元以内按10%提成奖金 else if (i=200000) bonus=bon1+(i-100000)
7、*0.075; /利润在10万元至20万时的奖金=400000) bonus=bon2+(i-200000)*0.05; /利润在20万元至40万时的奖金=600000) bonus=bon4+(i-400000)*0.03; /利润在40万元至60万时的奖金=1000000) bonus=bon6+(i-600000)*0.015; /利润在60万元至100万时的奖金 bonus=bon10+(i-1000000)*0.01; /利润在100万元以上时的奖金bonus=bonus10) c=10; switch(c) case 0: case 5: case 10:3.14题int t,a,
8、b,c,d;enter four numbers:cd;a=a, b=b, c=,d=d t=a;a=b;b=t; a=c; c=t;d) a=d; d=t; if (b t=b; b=c; b=d; t=c; c=d;the sorted sequence:, 3.15题int p,r,n,m,temp;please enter two positive integer numbers n,m:nm; if (nm) temp=n; n=m; m=temp; /把大数放在n中, 小数放在m中 p=n*m; /先将n和m的乘积保存在p中, 以便求最小公倍数时用 while (m!=0) /求n
9、和m的最大公约数 r=n%m; m=r;HCF=LCD=p/n & cZ letters+; else if (c= space+; else if (c09 digit+; other+;letter:letters, space:space, digit:digit, other:other 3.17题int a,n,i=1,sn=0,tn=0;a,n=:n; while (i=n) tn=tn+a; /赋值后的tn为i个a组成数的值 sn=sn+tn; /赋值后的sn为多项式前i项之和 a=a*10; +i;a+aa+aaa+.=sn3.18题float s=0,t=1; int n; for (n=1;=20;n+) t=t*n; / 求n! s=s+t; / 将各项累加 1!+2!+.+20!=3.19题int i,j,k,n;narcissus numbers are: for (n=100;1000; i=n/100; j=n/10-i*10; k=n%10; if (n = i*i*i + j*j*j + k*k*k) 3.20题 int main() const int m=1000; / 定义寻找范围 int k1,k2,k3,k4,k5,k6,k7,k8,k9,k10; int i,a,n,
copyright@ 2008-2022 冰豆网网站版权所有
经营许可证编号:鄂ICP备2022015515号-1