double类型的变量保留两位第三位四舍五入
问题描述
题目源于计算机二级C语言的程序设计题，做的时候有些懵，虽然说也是做出来了，但是代码写的实在不咋滴。
题目如下：输入一个double类型数据，保留两位，第三位四舍五入，由于题目需要返回值，所以不能直接打印。
1、标准答案
double h = 0.0;long t = 0, u = 0, v = 0;double w = 0;scanf_s("%lf \n", &h);t = h * 1000;//把第三位小数提到个位上u = t + 5;//四舍五入，如果大于五就进一位，没进位就不操作v = u / 10;//截掉第个位（第三位小数）w = (double)v / 100 ;//将剩余两位原小数位转化回小数printf("w=%lf \n", w);
最后一步w的值计算，拆开来写会出现没有小数位的情况。
结果：
2、非常好理解的办法（考试这么写的，也没问题，就是时间不大够）
double function(){int range = 0, GeWei = 0, ShiWei = 0, BaiWei = 0, QianWei = 0;//范围、个位、十位、百位、千位int SF = 0, BF = 0, QF = 0, point1 = 0, point2 = 0, point3 = 0;//SF：十分位、BF：百分位、WF：千分位//point1、2、3：小数点后三位int a[10] = {0 };double resultNum = 0.0, h = 0.0;scanf_s("%lf", &h);GeWei = (int)h % 10;//堆十取余获得各位resultNum = h * 0.1;//把十位拉到个位上ShiWei = (int)resultNum % 10;//获得身处个位的十位值resultNum = h * 0.01;//把百位拉到个位上BaiWei = (int)resultNum % 10;//获得身处个位的百位值resultNum = h * 0.001;//把千位拉到个位上QianWei = (int)resultNum % 10;//获得身处个位的千位值resultNum = h * 10;//把十分位扩到个位上SF = (int)resultNum % 10;//获得身处个位的十分位resultNum = h * 100;//把十分位扩到个位上BF = (int)resultNum % 10;//获得身处个位的百分位resultNum = h * 1000;//把十分位扩到个位上QF = (int)resultNum % 10;//获得身处个位的千分位printf("G = %d ", GeWei);printf("S = %d ", ShiWei);printf("B = %d ", BaiWei);printf("Q = %d ", QianWei);printf("\n");printf("SF = %d ", SF);printf("BF = %d ", BF);printf("QF = %d ", QF);printf("\n");a[range] = QianWei;//把他们都写进一个数组里方便后序操作a[++range] = BaiWei;a[++range] = ShiWei;a[++range] = GeWei;a[++range] = SF;point1 = range;//三个point记录分位在数组中的下标a[++range] = BF;point2 = range;a[++range] = QF;point3 = range;for (int i = 0; i <= range; i++){printf("a[%d] = %d ", i, a[i]);}if (a[point3] >= 5){printf("入\n");a[point2] += 1;a[point3] = 0;//四舍五入}h = QianWei * 1000 + BaiWei * 100 + ShiWei * 10 + GeWei + SF * 0.1 + a[point2] * 0.01;//按权重求和输出printf("h = %lf", h);return h;}
