87 lines
2.7 KiB
C
87 lines
2.7 KiB
C
// 计算含运算符 '&'（取两数较小值）与 '|'（取两数较大值）的表达式
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// 两者同优先级，按从左到右结合；遇到括号优先计算括号内表达式。
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// 输入：一行字符串，仅由数字、&、|、(、) 组成
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// 输出：一个整数，表示表达式的值
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#include <stdio.h>
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#include <stdlib.h>
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#include <string.h>
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#include <ctype.h>
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static inline int min_i(int a, int b) { return a < b ? a : b; }
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static inline int max_i(int a, int b) { return a > b ? a : b; }
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static inline void apply_top(int *vals, int *vtop, char *ops, int *otop) {
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char op = ops[--(*otop)];
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int b = vals[--(*vtop)];
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int a = vals[--(*vtop)];
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int r = (op == '&') ? min_i(a, b) : max_i(a, b);
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vals[(*vtop)++] = r;
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}
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int main(void) {
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// 读取整行输入
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char *line = NULL;
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size_t cap = 0;
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{
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// getline 若不可用则用 fgets 简易读取
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// Windows 平台无 getline，这里使用 fgets
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size_t bufcap = 16384; // 大于题目约束的 10000
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line = (char*)malloc(bufcap);
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if (!line) return 1;
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if (!fgets(line, (int)bufcap, stdin)) return 0;
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size_t len = strlen(line);
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// 去掉行尾 CR/LF
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while (len && (line[len-1] == '\n' || line[len-1] == '\r')) line[--len] = '\0';
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cap = len + 1;
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}
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size_t L = strlen(line);
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int *vals = (int*)malloc((L + 1) * sizeof(int));
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char *ops = (char*)malloc(L + 1);
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if (!vals || !ops) return 1;
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int vtop = 0, otop = 0;
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for (size_t i = 0; i < L;) {
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char c = line[i];
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if (isspace((unsigned char)c)) { ++i; continue; } // 忽略空白
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if (c >= '0' && c <= '9') {
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int x = 0;
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while (i < L && line[i] >= '0' && line[i] <= '9') {
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x = x * 10 + (line[i] - '0');
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++i;
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}
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vals[vtop++] = x;
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} else if (c == '&' || c == '|') {
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// 同优先级：在遇到 '(' 之前，应用所有已入栈的运算符
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while (otop > 0 && ops[otop - 1] != '(') {
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apply_top(vals, &vtop, ops, &otop);
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}
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ops[otop++] = c;
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++i;
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} else if (c == '(') {
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ops[otop++] = c;
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++i;
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} else if (c == ')') {
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while (otop > 0 && ops[otop - 1] != '(') {
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apply_top(vals, &vtop, ops, &otop);
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}
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if (otop > 0 && ops[otop - 1] == '(') --otop; // 弹出左括号
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++i;
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} else {
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// 非法字符：跳过以增强鲁棒性
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++i;
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}
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}
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while (otop > 0) {
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apply_top(vals, &vtop, ops, &otop);
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}
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if (vtop > 0) printf("%d\n", vals[vtop - 1]);
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free(vals);
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free(ops);
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free(line);
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return 0;
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} |