模拟/字符串处理 UVALive 6833 Miscalculatio

 

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 1 /*  2  模拟/字符串处理：主要是对*的处理，先把乘的预处理后再用加法，比如说是：1+2*3+4 = 1+..6+4 = 11  3 */  4 #include <cstdio>  5 #include <algorithm>  6 #include <cstring>  7 #include <string>  8 #include <cmath>  9 #include <vector>  10 #include <map>  11 #include <queue>  12 using namespace std;  13  14 typedef long long ll;  15 const int MAXN = 1e2 + 10;  16 const int INF = 0x3f3f3f3f;  17 char s[MAXN];  18 int len;  19  20 void solve(void)  21 {  22 int p1 = 0, p2 = 0;  23 for (int i=1; i<=len; ++i)  24  {  25 if (s[i] == '*')  26  {  27 p1 = p2 = i;  28 while (p1 >= 1 && s[p1] != '+') p1--; p1++;  29 while (p2 <= len && s[p2] != '+') p2++; p2--;  30 ll tmp = 0;  31 for (int j=p1; j<=p2; j+=2)  32  {  33 if (j == p1) tmp = s[j] - '0';  34 else tmp = tmp * (s[j] - '0');  35  }  36 // printf ("p1: %d p2: %d\ntmp: %d\n", p1, p2, tmp);  37 int p = p2;  38 while (tmp)  39  {  40 s[p--] = tmp % 10 + '0';  41 tmp /= 10;  42  }  43 for (int j=p1; j<=p; ++j) s[j] = '.';  44  }  45  }  46 }  47  48 int main(void) //UVALive 6833 Miscalculation  49 {  50 // freopen ("B.in", "r", stdin);  51  52 while (scanf ("%s", s + 1) == 1)  53  {  54 ll y; scanf ("%lld", &y);  55 ll sum_l = 0, sum_m = 0;  56  57 len = strlen (s + 1); char op; int x = 0;  58 for (int i=1; i<=len; ++i)  59  {  60 if (i & 1)  61  {  62 if (i == 1) sum_l = s[i] - '0';  63 else  64  {  65 if (op == '+') sum_l += s[i] - '0';  66 else sum_l = sum_l * (s[i] - '0');  67  }  68  }  69 else op = s[i];  70  }  71  72  solve ();  73 for (int i=1; i<=len; ++i)  74  {  75 if (s[i] == '.' || s[i] == '+') continue;  76 else if (s[i] >= '0' && s[i] <= '9')  77  {  78 ll res = 0; int j;  79 for (j=i; j<=len; ++j)  80  {  81 if (s[j] == '+') break;  82 res = res * 10 + (s[j] - '0');  83  }  84 sum_m += res;  85 i = j - 1;  86  }  87  }  88  89 if (y == sum_l && y == sum_m) puts ("U");  90 else if (y != sum_l && y != sum_m) puts ("I");  91 else  92  {  93 if (y == sum_l) puts ("L");  94 else puts ("M");  95  }  96  97 // printf ("%s\n", s + 1);  98 // printf ("%lld %lld \n", sum_l, sum_m);  99  } 100 101 return 0; 102 }