UVA10700

来源：互联网发布：淘宝ted baker 编辑：程序博客网时间：2024/05/16 15:39

题意：求出一组数据可能出现的最小值和最大值，符号只有‘+’和‘*’

思路：通常我们遵循乘法优先，可以求出最小值，至于最大值，我们就让加法成为优先级，相当于，先算加法，再算乘法，主要用数组模拟

#include <stdio.h>#include <string.h>#define N 30char str[N];double s[N], max, min;int main() {    int n;    scanf("%d", &n);    getchar();    while (n--) {        gets(str);        int i, top;        char ch;        i = top = 0;        ch = '+';        while (str[i] != '\0') {            int num = 0;            while (str[i] >= '0' && str[i] <= '9') {                num = num * 10 + str[i] - '0';                i++;             }            if (ch == '+')                s[++top] = num;            else                s[top] *= num;            if (str[i] != '\0')                ch = str[i++];        }        min = 0;        for (int i = 1; i <= top; i++)            min += s[i];        i = top = 0;        ch = '*';         while (str[i] != '\0') {            int num = 0;            while (str[i] >= '0' && str[i] <= '9') {                num = num * 10 + str[i] - '0';                i++;             }            if (ch == '*')                s[++top] = num;            else                s[top] += num;            if (str[i] != '\0')                ch = str[i++];         }        max = 1;        for (int i = 1; i <= top; i++)            max *= s[i];        printf("The maximum and minimum are %.0lf and %.0lf.\n", max, min);    }    return 0;}