C语言编程题

1将程序段填充完整（实现12个月每个月天数的输出）

case 1: case 3: case 5: case 7: case 8: case 10: case 12: days=31;break;

case 4: case 6: case 9: case 11: days=30;break;

case 2:

if((yy%4==0 && yy%100!=0) || yy%400==0) days=29;

else days=28;

break;

2输入4个整数a,b,c,d,编写程序，将它们按从大到小的顺序输出（if语句实现）

int a,b,c,d;

scanf("%d %d %d %d",&a,&b,&c,&d);

if(a>b && a>c && a>d) printf("%d ",a);

if(b>a && b>c && b>d) printf("%d ",b);

if(c>a && c>b && c>d) printf("%d ",c);

if(d>a && d>b && d>c) printf("%d ",d);

3假设奖金税率如下（ma代表税前奖金且ma>0,r代表税率），利用switch语句编写程序对输入的一个奖金数，输出税率和应交税款及实得奖金数（扣除奖金税后）。

int ma;

float r,ts,sd;

scanf("%d",&ma);

switch(ma/1000)

{

case 0: r=0;break;

case 1: r=005;break;

case 2: r=008;break;

default: r=0

#include "stdioh"

#include "conioh"

main()

{float i,b,b1,b2,b4,b6,b10;

printf("Input Please:");

scanf("%f",&i);

b1=10000001;

b2=b1+1000000075;

b4=b2+200000005;

b6=b4+200000003;

b10=b6+4000000015;

if (100000>i)

b=i01;

else

if(100000<i&&i<=200000)//程序中这样表示！下面一样！

b=b1+(i-100000)0075;

else

if(200000<i&&i<=400000)

b=b2+(i-200000)005;

else

if(400000<i&&i<=600000)

b=b4+(i-400000)003;

else

if(600000<i&&i<=1000000)

b=b6+(i-600000)0015;

else

b=b10+(i-1000000)001;

printf("Bonus=%62f",b);

getch();

}

program Poor_Xed;

var

a, b : array of longint;

x : array of record

above : array of longint;

below, money : longint

end;

n, m, i, count : longint;

procedure recur(k : longint);

var

i : longint;

begin

dec(count);

with x[k] do

for i:=1 to above[0] do begin

if money >= x[above[i]]money then

x[above[i]]money := money + 1;

dec(x[above[i]]below);

if x[above[i]]below = 0 then

recur(above[i])

end

end;

begin

readln(n, m);

setlength(x, n+1);

setlength(a, m+1);

setlength(b, m+1);

for i:=1 to m do

readln(a[i], b[i]);

for i:=1 to n do

with x[i] do begin

setlength(above, 1);

above[0] := 0;

end;

for i:=1 to m do

with x[b[i]] do begin

inc(above[0]);

if above[0] >= length(above) then

setlength(above, length(above) 2);

above[above[0]] := a[i]

end;

for i:=1 to n do

x[i]below := 0;

for i:=1 to m do

inc(x[a[i]]below);

for i:=1 to n do

x[i]money := 100;

count := n;

for i:=1 to n do

if x[i]below = 0 then

recur(i);

if count = 0 then begin

for i:=1 to n do {begin

writeln(x[i]money);}

count := count + x[i]money;

{end;

writeln;}

writeln(count)

end else

writeln('Poor Xed')

end

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