本文共 2063 字，大约阅读时间需要 6 分钟。
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 5 #include 6 #include 7 #include 8 #include 9 #include 10 #include 11 #include 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 }
转载于:https://www.cnblogs.com/Running-Time/p/4558848.html