C语言程序设计之循环结构.docxVIP

四、循环结构1、求1-1/4+1/7-1/10+1/13-1/16+……程序填空，不要改变与输入输出有关的语句。输入一个正整数repeat (0repeat10)，做repeat次下列运算：读入1个正实数 eps，计算并输出下式的值，精确到最后一项的绝对值小于 eps(保留6位小数)。请使用 while 语句实现循环。输入输出示例：括号内是说明输入2 (repeat=2)1E-4 (eps=1E-4)0.1 (eps=0.1)输出sum = 0.835699sum = 0.869780#include stdio.h#include math.hint main(void){ int denominator, flag; int repeat, ri; double eps, item, sum; scanf(%d, repeat); for(ri = 1; ri = repeat; ri++){ scanf(%le, eps); sum=0; denominator =1;flag=1;item=1; while(fabs(item)=eps){ item= flag*1.0/ denominator; sum=sum+item;denominator = denominator +3;flag=-flag;} printf(sum = %.6f\n, sum); }}2、求奇数和程序填空，不要改变与输入输出有关的语句。输入一个正整数repeat (0repeat10)，做repeat次下列运算：读入一批正整数(以零或负数为结束标志)，求其中的奇数和。请使用while语句实现循环。输入输出示例：括号内是说明输入2(repeat=2)1 3 90 7 08 7 4 3 70 5 6 101 -1输出The sum of the odd numbers is 11.The sum of the odd numbers is 116.#include stdio.hint main(void){ int x, sum; int repeat, ri; scanf(%d, repeat); for(ri = 1; ri = repeat; ri++){ scanf(%d, x); sum=0; while(x0) { if(x%2) sum=sum+x; scanf(%d, x); } printf(The sum of the odd numbers is %d.\n, sum); }}3、求整数的位数程序填空，不要改变与输入输出有关的语句。输入一个正整数 repeat (0repeat10)，做 repeat 次下列运算：输入一个整数 in，求它的位数。例如123的位数是3。请使用do-while语句实现循环。输入输出示例：括号内是说明输入4 (repeat=4) 12345 (in=12345) -100 (in=-100) -1 (in=-1) 1290 (in=1290) 输出count = 5 (12345的位数是5)count = 3 (-100的位数是3)count = 1 (-1的位数是1)count = 4 (99的位数是2)#include stdio.hint main(void){ int count, in; int repeat, ri; scanf(%d, repeat); for(ri = 1; ri = repeat; ri++){ scanf(%d, in);count =0; if(in0) in=-in; do{ count ++; in=in/10; }while(in); printf(count = %d\n, count); }}4、求最小值程序填空，不要改变与输入输出有关的语句。输入一个正整数repeat (0repeat10)，做repeat次下列运算：输入一个正整数n, 再输入n个整数，输出最小值。输入输出示例：括号内是说明输入3 (repeat=3) 4 -2 -123 100 0 4 -9 -1 1 -8 3 5 3 1 输出min = -123min = -9min = 1#include stdio.hint main(void){ int i, min, n, x; int repeat, ri; scanf(%d, repeat); for(ri = 1; ri = repeat; ri++){ scanf(%d, n); scanf(%d, x);min=x; for(i=1;in;i++){ scanf(%d, x); if(xmin) min=x;} printf(min =