1034. 有理数四则运算(20)
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1034. 有理数四则运算(20)
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本題要求編寫程序,計算2個有理數的和、差、積、商。
輸入格式:
輸入在一行中按照“a1/b1 a2/b2”的格式給出兩個分數形式的有理數,其中分子和分母全是整型范圍內的整數,負號只可能出現在分子前,分母不為0。
輸出格式:
分別在4行中按照“有理數1 運算符 有理數2 = 結果”的格式順序輸出2個有理數的和、差、積、商。注意輸出的每個有理數必須是該有理數的最簡形式“k a/b”,其中k是整數部分,a/b是最簡分數部分;若為負數,則須加括號;若除法分母為0,則輸出“Inf”。題目保證正確的輸出中沒有超過整型范圍的整數。
輸入樣例1:
2/3 -4/2
輸出樣例1:
2/3 + (-2) = (-1 1/3)
2/3 - (-2) = 2 2/3
2/3 * (-2) = (-1 1/3)
2/3 / (-2) = (-1/3)
輸入樣例2:
5/3 0/6
輸出樣例2:
1 2/3 + 0 = 1 2/3
1 2/3 - 0 = 1 2/3
1 2/3 * 0 = 0
1 2/3 / 0 = Inf
錯誤代碼:
/*************************************************************************> File Name: 1034.c> Author: YueBo > Function: 有理數的四則運算> Created Time: 2016年11月18日 星期五 12時55分10秒************************************************************************/#include <stdio.h> #include <stdlib.h> #include <math.h> struct rNum1 {long long a, b; };void print(struct rNum1* pr) {long long m, m1, n, q, i;n = abs(pr->b);if (n == 0) { //如果分母為0,啥都不用說了printf("Inf");return;}m = abs(pr->a) % abs(pr->b);q = pr->a / pr->b;if (m != 0) { //化簡真分數for (i = 2; i <= m; i++) {if (m%i==0 && n%i==0) {m = m / i;n = n / i;i = 1; //剛開始i=2老是出錯}}}if (pr->a >= 0) {if (m == 0) { //如果分子為0printf("%lld", q);} else if (q != 0) {printf("%lld %lld/%lld", q, m, n);} else {printf("%lld/%lld", m, n);}} else {if (m == 0) {printf("(%lld)", q);}else if (q != 0) {printf("(%lld %lld/%lld)", q, m, n);} else {printf("(-%lld/%lld)", m, n);}} }void plus(struct rNum1* pr1, struct rNum1* pr2, struct rNum1* pr3) {pr3->b = pr1->b * pr2->b;pr3->a = pr1->a * pr2->b + pr2->a * pr1->b; }void minus(struct rNum1* pr1, struct rNum1* pr2, struct rNum1* pr3) {pr3->b = pr1->b * pr2->b;pr3->a = pr1->a * pr2->b - pr2->a * pr1->b; }void times(struct rNum1* pr1, struct rNum1* pr2, struct rNum1* pr3) {pr3->b = pr1->b * pr2->b;pr3->a = pr1->a * pr2->a; }void division(struct rNum1* pr1, struct rNum1* pr2, struct rNum1* pr3) {if (pr2->a > 0) {pr3->b = pr1->b * pr2->a;pr3->a = pr1->a * pr2->b;} else {pr3->b = (-1)*pr1->b * pr2->a;pr3->a = (-1)*pr1->a * pr2->b;} }int main() {struct rNum1 r1, r2, r3;scanf("%lld/%lld", &r1.a, &r1.b);scanf("%lld/%lld", &r2.a, &r2.b);//加法plus(&r1, &r2, &r3);print(&r1);printf(" + ");print(&r2);printf(" = ");print(&r3);printf("\n");//減法minus(&r1, &r2, &r3);print(&r1);printf(" - ");print(&r2);printf(" = ");print(&r3);printf("\n");//乘法times(&r1, &r2, &r3);print(&r1);printf(" * ");print(&r2);printf(" = ");print(&r3);printf("\n");//除法division(&r1, &r2, &r3);print(&r1);printf(" / ");print(&r2);printf(" = ");print(&r3);return 0; }提示錯誤:
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