C语言cfree题目.docx

2417Description输入某年某月某日，判断这一天是这一年的第几天？Input输入为一行输入格式为YYYY-MM-DDOutput输出这天是这一年的第几天Sample Input2007-01-01Sample Output1SourceYhr#includeint main ()int a,b,c,d,f;scanf (%d-%d-%d,&a,&b,&c);if (a%400 = 0)f = 1;else if (a%4 = 0 & a%100 != 0)f = 1 ;elsef = 0 ;switch(b) case 1: d = 0 ;break; case 2: d = 31 ;break; case 3: d = 31+28 ;break; case 4: d = 31+28+31;break; case 5: d = 31+28+31+30;break; case 6: d = 31+28+31+30+31;break; case 7: d = 31+28+31+30+31+30;break; case 8: d = 31+28+31+30+31+30+31 ;break; case 9: d = 31+28+31+30+31+30+31+31;break; case 10: d = 31+28+31+30+31+30+31+31+30 ;break; case 11: d = 31+28+31+30+31+30+31+31+30+31 ;break; case 12: d = 31+28+31+30+31+30+31+31+30+31+30;break; default: printf (input error!);break;d = d +c ;if (b 2)d = d + f*1 ;printf(%dn,d);return 0;Description输入数字1-7，打印出对应的英文（Monday-Sunday）. Input输入数字1-73502Output输出该数字对应的英文单词Sample Input7Sample OutputSundaySourcejsj_5_3#includeint main ()int n;scanf (%d,&n);switch(n) case 1: printf (Mondayn) ;break; case 2: printf (Tuesdayn) ;break; case 3: printf (Wednesdayn) ;break; case 4: printf (Thursdayn);break; case 5: printf (Fridayn);break; case 6: printf (Saturdayn);break; case 7: printf (Sundayn);break; default: printf (input error!);break;return 0;3503Description输入一个百分制成绩，转换为五级记分制成绩并输出。Input输入一个score（0=score good score:80-89 - nicer score:70-79 - medium score:60-69 - pass score:0-59 - fail Sample Input67Sample OutputpassSourcemgc1008#includeint main ()int s;scanf (%d,&s);if (s = 90 & s = 80 & s = 70 & s = 60 & s = 69) printf (passn); else printf (failn);return 0;3504要求根据以下函数关系，对输入的任意整数x，计算出相应的y值。要求用switch语句实现。x的值 y的值x0 y=0 0=x10 y=x 10=x=20 y=30 Input输入的任意整数x Output计算出相应的y值Sample Input39Sample Outputy=30.000000Sourcejsj_4_4#includeint main ()int x,y,t;scanf (%d,&x); if (x 0) t = 1; else if (x = 0) t= 2; else if (x = 10) t = 3; else t = 4; switch(t) case 1: printf (y=0n) ;break; case 2: printf (y=%dn,x) ;break; case 3: printf (y=10n) ;break; case 4: printf (y=30n);break; default: printf (input error!);break;return 0;3505Total Submit:526 Accepted:268 Description输入一个实数x(float)，计算并输出下列分段函数的值。请调用数学函数计算。 y=x5+2x+1（x0） y=ex-lg(x)（x0）Input输入一个实数x。Output根据计算公式计算y值并输出（保留3位小数)。Sample Input1.2Sample Outputy=3.241Hint求ex的函数：exp(x) 求xy的函数：pow(x,y) 求lg(x)的函数：log10(x) Sourcemgc10113813#include#includeint main ()float x,y;scanf (%f,&x);if (x = 0)y = pow(x,5) + 2*x + 1;elsey = exp(x) - log10(x) ;printf (%.3fn,y);return 0 ;Description输入1个3位数，分离出它的百位、十位和个位，反转后输出Input输入1个3位整数Output输出3位整数的反转数Sample Input250Sample Output052Hint分离出各位数字可以用取余和除数注意在C语言里，2个整数相乘除结果还是整数比如8/3在C语言里结果是2 取余采用符号% 比如8%3的结果应该是2即8除以3后的余数Source3814#includeint main ()int x,a,b,c;scanf(%d,&x);a = x/100;b = (x/10)%10);c = x%10;printf (%d%d%dn,c,b,a);return 0 ;Description输入2个整数，交换后输出Input输入2个整数Output交换后输出Sample Input825 23Sample Output23 825Hint#includeint main ()int a,b,c;scanf (%d%d,&a,&b);c = a;a = b;b = c;printf (%d %dn,a,b);return 0 ;在C语言里，交换可以用3变量法，要实现a和b交换，先将变量a的值保存到变量c,然后将变量b的值赋值给a,然后将c的值（最初a的值）赋值给b 赋值语句c=a; 等号左边是一个变量，用来存放值，右边是一个表达式，计算这个表达式后保存到等号左边的变量中Source3815Description鸡和兔关在同一笼子中，已知总数量n和腿的总数量m,求鸡的数量和兔的数量。Input输入2个整数分别是n和mOutput输出鸡的数量和兔的数量，中间用空格隔开，如果无解则输出NoSample Input14 32Sample Output12 2Source3816Description输入2个整数，按照从大到小输出Input输入2个整数分别是n和mOutput按照从大到小输出，中间用空格隔开Sample Input14 32Sample Output32 14Source3817#includeint main ()int x,a,b,c;scanf(%d,&x);a = x/100;b = (x/10)%10);c = x%10;printf (%d%d%dn,c,b,a);return 0 ;Description输入3个整数，将它们从大到小输出Input输入3个整数，Output从大到小输出，中间用空格隔开Sample Input2 5 1Sample Output5 2 1Source3826Description输入三角形的三条边，判断是否是直角三角形Input输入三角形3条边(均为整数）Output如果3条边组成的是直角三角形，输出yes否则输出noSample Input3 4 5Sample OutputYes#includeint main ()int a,b,c;scanf (%d%d%d,&a,&b,&c);if (a*a + b*b = c*c | a*a + c*c = b*b | b*b + c*c = a*a)printf (yesn);elseprintf (non);return 0 ;3827Description输入年份，判断是否闰年Input输入一个整数n表示年份Output如果是闰年，输出yes,否则输出noSample Input2000Sample OutputyesSource3828Description输入年份，判断是否闰年Input输入一个整数n表示年份Output如果是闰年，输出yes,否则输出noSample Input2000Sample OutputyesSource3829Description每年我们都有不少同学去参加程序设计竞赛。考虑到参赛名额的限制，我们的教练制定了以下规则：（1）如果等级分排名前6或者训练缺席次数不多于2次的队伍允许代表学校参赛（2）如果等级分排名前20且训练缺席次数小于5次的允许参赛Inp