ZOJ-2021

来源：互联网发布：朱元璋厉害知乎婚姻编辑：程序博客网时间：2024/04/26 06:18

繁琐的括号消除，比1423那道复杂多了，首先要理清全部的可消除情况的逻辑，然后慢慢写。。我的处理方法是先算出来左右括号的下标对应关系，然后分括号前后的符号和括号里的符号各种情况来判断，注意，统计括号内符号的时候要跳过那些包含在子括号内的符号，比如说((a*b)+(c/d))，那么统计的时候应该只算一个加号。本题数据量超大，逻辑有一点错的话肯定过不了。。把数据弄过来，错了就慢慢改吧。。我也是一点点改对的

简化分支后的代码:

#include<stdio.h>#include<string.h>#include<ctype.h>static char prev(char *s, int index){int i;for (i = index - 1; i >= 0 && s[i] != '('; i--)if (s[i] != ' ')return s[i];return -1;}static char next(char *s, int index){int i;for (i = index + 1; s[i] != '\0' && s[i] != ')'; i++)if (s[i] != ' ')return s[i];return -1;}static void count(char *s, int begin, int end, int *par, int *map){memset(map, 0, 4 * sizeof(int));int i;for (i = begin + 1; i < end; i++){if (s[i] == '(')i = par[i];if (s[i] == '+')map[0] = 1;if (s[i] == '-')map[1] = 1;if (s[i] == '*')map[2] = 1;if (s[i] == '/')map[3] = 1;}}static void removep(char *s, int left, int right){s[left] = ' ';s[right] = ' ';}int main(){int t, par[256], stack[100], map[4], index;scanf("%d", &t);getchar();char s[256];while (t--){gets(s);memset(par, 0, sizeof(par));index = 0;int i;for (i = 0; s[i] != '\0'; i++)if (s[i] == '(')stack[index++] = i;else if (s[i] == ')')par[stack[--index]] = i;for (i = 0; i < 256; i++)if (par[i]){char pp = prev(s, i);char nn = next(s, par[i]);count(s, i, par[i], par, map);if (pp == -1 || pp == '+'){if (nn == -1 || nn == '+' || nn == '-')removep(s, i, par[i]);else if (!map[0] && !map[1])removep(s, i, par[i]);}else if ((pp == '*' || pp == '-') && !map[0] && !map[1])removep(s, i, par[i]);else if (pp == '/' && !map[0] && !map[1] && !map[2] && !map[3])removep(s, i, par[i]);}for (i = 0; s[i] != '\0'; i++)if (s[i] != ' ')putchar(s[i]);putchar('\n');}return 0;}

原始代码：

#include<stdio.h>#include<string.h>#include<ctype.h>static char prev(char *s, int index){int i;for (i = index - 1; i >= 0 && s[i] != '('; i--)if (s[i] != ' ')return s[i];return -1;}static char next(char *s, int index){int i;for (i = index + 1; s[i] != '\0' && s[i] != ')'; i++)if (s[i] != ' ')return s[i];return -1;}static void count(char *s, int begin, int end, int *par, int *map){memset(map, 0, 4 * sizeof(int));int i;for (i = begin + 1; i < end; i++){if (s[i] == '(')i = par[i];if (s[i] == '+')map[0] = 1;if (s[i] == '-')map[1] = 1;if (s[i] == '*')map[2] = 1;if (s[i] == '/')map[3] = 1;}}static void process(char *s){int i;for (i = 0; s[i] != '\0'; i++)if (islower(s[i])){int j = i - 1;while (j >= 0 && s[j] == '(')if (s[2 * i - j] == ')'){s[j] = ' ';s[2 * i - j] = ' ';j--;}elsebreak;}}int main(){int t, par[256], stack[100], map[4], index;scanf("%d", &t);getchar();char s[256];while (t--){gets(s);process(s);memset(par, 0, sizeof(par));index = 0;int i;for (i = 0; s[i] != '\0'; i++)if (s[i] == '(')stack[index++] = i;else if (s[i] == ')')par[stack[--index]] = i;for (i = 0; i < 256; i++)if (par[i]){char pp = prev(s, i);char nn = next(s, par[i]);count(s, i, par[i], par, map);if (pp == -1){if (nn == -1 || nn == '+' || nn == '-'){s[i] = ' ';s[par[i]] = ' ';}else{if (!map[0] && !map[1]){s[i] = ' ';s[par[i]] = ' ';}}}else if (pp == '*'){if (!map[0] && !map[1]){s[i] = ' ';s[par[i]] = ' ';}}else if (pp == '/'){if (!map[0] && !map[1] && !map[2] && !map[3]){s[i] = ' ';s[par[i]] = ' ';}}else if (pp == '+'){if (nn == -1 || nn == '+' || nn == '-'){s[i] = ' ';s[par[i]] = ' ';}else if (!map[0] && !map[1]){s[i] = ' ';s[par[i]] = ' ';}}else if (pp == '-'){if (!map[0] && !map[1]){s[i] = ' ';s[par[i]] = ' ';}}}for (i = 0; s[i] != '\0'; i++)if (s[i] != ' ')putchar(s[i]);putchar('\n');}return 0;}

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