080425
来源：互联网 发布：淘宝上有卖电棍的吗 编辑：程序博客网 时间：2024/04/29 15:58
编写一个程序，要求计算出生日到计算日的总天数
算法：
1)输入出生年year1、月month1、日date1和计算日年year2、月month2、日date2。
2)统计year1到year2-1的总天数：
for (n=year1,days=0;n<year2;n++)
if (n是闰年) days+=366
else days+=365;
3)计算出出生日和计算日是当年的第days1和days2。
4)总天数等于days+days2-days1
根据结果，判断体力、情绪和智力状态
方法如下：
体力状态：用总天数除以23，余数在0~11为体力高潮期，否则为体力低潮期。
情绪状态：用总天数除以28，余数在0~14为情绪高潮期，否则为情绪低潮期。
智力状态：用总天数除以33，余数在0~16为智力高潮期，否则为智力低潮期。
#include <stdio.h>
void main(){
int year1,year2;
int month1,month2;
int date1,date2;
int n,m;
int days = 0;
int days1 = 0;
int days2 = 0;
int all_days;
printf("please input your brithday(for example:year month day):");
scanf("%d%d%d",&year1,&month1,&date1);
printf("please input today(for example:year month day):");
scanf("%d%d%d",&year2,&month2,&date2);
for(n = year1; n < year2; n++){
if((((n % 4) == 0)&&((n % 100) != 0)) || (n % 400) == 0){
days = days + 366;
}
else
days = days + 365;
}
printf("the year1to year2 has %d days/n",days);
switch(month1){
case 1:
days1=date1;
printf("the day is %d days in the year1/n",days1);
break;
case 2:
days1=date1+31;
printf("the day is %d days in the year1/n",days1);break;
case 3:
if((((n % 4) == 0)&&((n % 100) != 0)) || (n % 400) == 0){
days1=date1+31+29;
printf("the day is %d days in the year1/n",days1);
break;
}
else {
days1=date1+31+28;
printf("the day is %d days in the year1/n",days1);
break;
}
case 4:
if((((n % 4) == 0)&&((n % 100) != 0)) || (n % 400) == 0){
days1=date1+31+29+31;
printf("the day is %d days in the year1/n",days1);
break;
}
else {
days1=date1+31+28+31;
printf("the day is %d days in the year1/n",days1);
break;
}
case 5:
if((((n % 4) == 0)&&((n % 100) != 0)) || (n % 400) == 0){
days1=date1+31+29+31+30;
printf("the day is %d days in the year1/n",days1);
break;
}
else{
days1=date1+31+28+31+30;
printf("the day is %d days in the year1/n",days1);
break;
}
case 6:
if((((n % 4) == 0)&&((n % 100) != 0)) || (n % 400) == 0){
days1=date1+31+29+31+30+31;
printf("the day is %d days in the year1/n",days1);
break;
}
else {
days1=date1+31+28+31+30+31;
printf("the day is %d days in the year1/n",days1);
break;
}
case 7:
if((((n % 4) == 0)&&((n % 100) != 0)) || (n % 400) == 0){
days1=date1+31+29+31+30+31+30;
printf("the day is %d days in the year1/n",days1);
break;
}
else{
days1=date1+31+28+31+30+31+30;
printf("the day is %d days in the year1/n",days1);
break;
}
case 8:
if((((n % 4) == 0)&&((n % 100) != 0)) || (n % 400) == 0){
days1=date1+31+29+31+30+31+30+31;
printf("the day is %d days in the year1/n",days1);
break;
}
else{
days1=date1+31+28+31+30+31+30+31;
printf("the day is %d days in the year1/n",days1);
break;
}
case 9:
if((((n % 4) == 0)&&((n % 100) != 0)) || (n % 400) == 0){
days1=date1+31+29+31+30+31+30+31+31;
printf("the day is %d days in the year1/n",days1);
break;
}
else {
days1=date1+31+28+31+30+31+30+31+31;
printf("the day is %d days in the year1/n",days1);
break;
}
case 10:
if((((n % 4) == 0)&&((n % 100) != 0)) || (n % 400) == 0){
days1=date1+31+29+31+30+31+30+31+31+30;
printf("the day is %d days in the year1/n",days1);
break;
}
else{
days1=date1+31+28+31+30+31+30+31+31+30;
printf("the day is %d days in the year1/n",days1);
break;
}
case 11:
if((((n % 4) == 0)&&((n % 100) != 0)) || (n % 400) == 0){
days1=date1+31+29+31+30+31+30+31+31+30+31;
printf("the day is %d days in the year1/n",days1);
break;
}
else {
days1=date1+31+28+31+30+31+30+31+31+30+31;
printf("the day is %d days in the year1/n",days1);
break;
}
case 12:
if((((n % 4) == 0)&&((n % 100) != 0)) || (n % 400) == 0){
days1=366-31+date1;
printf("the day is %d days in the year1/n",days1);
break;
}
else {
days1=365-31+date1;
printf("the day is %d days in the year1/n",days1);
break;
}
default :
printf("data error");
}
switch(month2){
case 1:
days2=date2;
printf("the day is %d days in the year2/n",days2);
break;
case 2:
days2=date2+31;
printf("the day is %d days in the year2/n",days2);break;
case 3:
if((((n % 4) == 0)&&((n % 100) != 0)) || (n % 400) == 0){
days2=date2+31+29;
printf("the day is %d days in the year2/n",days2);
break;
}
else {
days2=date2+31+28;
printf("the day is %d days in the year2/n",days2);
break;
}
case 4:
if((((n % 4) == 0)&&((n % 100) != 0)) || (n % 400) == 0){
days2=date2+31+29+31;
printf("the day is %d days in the year2 /n",days2);
break;
}
else {
days2=date2+31+28+31;
printf("the day is %d days in the year2 /n",days2);
break;
}
case 5:
if((((n % 4) == 0)&&((n % 100) != 0)) || (n % 400) == 0){
days2=date2+31+29+31+30;
printf("the day is %d days in the year2 /n",days2);
break;
}
else{
days2=date2+31+28+31+30;
printf("the day is %d days in the year2 /n",days2);
break;
}
case 6:
if((((n % 4) == 0)&&((n % 100) != 0)) || (n % 400) == 0){
days2=date2+31+29+31+30+31;
printf("the day is %d days in the year2 /n",days2);
break;
}
else {
days2=date2+31+28+31+30+31;
printf("the day is %d days in the year2/n",days2);
break;
}
case 7:
if((((n % 4) == 0)&&((n % 100) != 0)) || (n % 400) == 0){
days2=date2+31+29+31+30+31+30;
printf("the day is %d days in the year2 /n",days2);
break;
}
else{
days2=date2+31+28+31+30+31+30;
printf("the day is %d days in the year2 /n",days2);
break;
}
case 8:
if((((n % 4) == 0)&&((n % 100) != 0)) || (n % 400) == 0){
days2=date2+31+29+31+30+31+30+31;
printf("the day is %d days in the year2 /n",days2);
break;
}
else{
days2=date2+31+28+31+30+31+30+31;
printf("the day is %d days in the year2 /n",days2);
break;
}
case 9:
if((((n % 4) == 0)&&((n % 100) != 0)) || (n % 400) == 0){
days2=date2+31+29+31+30+31+30+31+31;
printf("the day is %d days in the year2 /n",days2);
break;
}
else {
days2=date2+31+28+31+30+31+30+31+31;
printf("the day is %d days in the year2 /n",days2);
break;
}
case 10:
if((((n % 4) == 0)&&((n % 100) != 0)) || (n % 400) == 0){
days2=date2+31+29+31+30+31+30+31+31+30;
printf("the day is %d days in the year2/n",days2);
break;
}
else{
days2=date2+31+28+31+30+31+30+31+31+30;
printf("the day is %d days in the year2/n",days2);
break;
}
case 11:
if((((n % 4) == 0)&&((n % 100) != 0)) || (n % 400) == 0){
days2=date2+31+29+31+30+31+30+31+31+30+31;
printf("the day is %d days in the year2/n",days2);
break;
}
else {
days2=date2+31+28+31+30+31+30+31+31+30+31;
printf("the day is %d days in the year2 /n",days2);
break;
}
case 12:
if((((n % 4) == 0)&&((n % 100) != 0)) || (n % 400) == 0){
days2=366-31+date2;
printf("the day is %d days in the year2 /n",days2);
break;
}
else {
days2=365-31+date2;
printf("the day is %d days in the year2 /n",days2);
break;
}
default :
printf("data error/n");
}
all_days=days+days2-days1;
printf("in all days is %d/n",all_days);
if ((all_days/23)>=0&&(all_days/23)<=11){
printf("sinew is climactic ! /n");
}
else
printf("sinew is lower!/n");
if ((all_days/28)>=0&&(all_days/28)<=41){
printf("mood is climactic !/n ");
}
else
printf("mood is lower!/n");
if ((all_days/33)>=0&&(all_days/33)<=16){
printf("wit is climactic ! /n");
}
else
printf("wit is lower!/n");
}
1. 编写一个程序求c =a+ | b | , 输出c 的值, 其中a和b为整数
( 绝对值函数请查函数库中的math.h 文件)
(1)解
#include <stdio.h>
int main(){
float a,b;
float sum;
scanf("%f%f",&a,&b);
if(b < 0){
b = -b;
}
sum = a + b;
printf("%f/n",sum);
}
（2）解
#include <stdio.h>
#include <math.h>
void main(){
float a,b;
float c;
float sum;
scanf("%f%f",&a,&b);
c=abs(b);
sum=a+c;
printf("sum is %f/n",sum);
}
2. 写出下面程序的输出结果
#include <stdio.h>
void main( )
{ int x , y , z ;
x = -3+4*5-6 ; printf("x=%d/n", x);//x=11
y = -3*4%-6/5 ; printf("y=%d/n", y);//y=0
z = (7+6)%5/2 ; printf("z=%d/n", z);//z=1
x = 3 ;
y = ++x -1; printf("x=%d,y=%d/n", x , y);//x=4,y=3
z = x--+1 ; printf("x=%d,z=%d/n", x , z);//x=3,z=5
}
3. 求出下列表达式的值
假设 x = 3, y = 4, z = 4
(1) (z >= y >= x)? 1 : 0
(2) z>=y && y>=x
假设 x = 3, y = 2, z = 1
(3) x < y ? x++ : y++
(4) z +=(x < y? x++ : y++)
