P1310 表达式的值 （表达式计算）

Posted 2021-03-12 feiief

题目链接

 

 

 

 

解法：

栈

 1 #include <bits/stdc++.h>
 2 # define LL long long
 3 using namespace std;
 4 
 5 const int mod=10007;
 6 stack<char> ope;
 7 stack<int> one;
 8 stack<int> zero;
 9 
10 void calc(){
11     char o=ope.top();
12     ope.pop();
13     int one1=one.top();
14     one.pop();
15     int one2=one.top();
16     one.pop();
17     int zero1=zero.top();
18     zero.pop();
19     int zero2=zero.top();
20     zero.pop();
21     int t1;
22     int t0;
23     if(o==‘*‘){
24         t1=(one1*one2)%mod;
25         t0=((one1*zero2)%mod+(one2*zero1)%mod+(zero1*zero2)%mod)%mod;
26     }else {
27         t1=((zero1*one2)%mod+(one1*zero2)%mod+(one2*one1)%mod)%mod;
28         t0=(zero1*zero2)%mod;
29     }
30     one.push(t1);
31     zero.push(t0);
32 }
33 
34 int main(){
35     int L;
36     cin>>L;
37     string s;
38     cin>>s;
39     s=‘(‘+s+‘)‘;
40 
41     for(int i=0;i<s.size();++i){
42         if(s[i]==‘(‘){
43             ope.push(s[i]);
44             if(s[i+1]!=‘(‘){
45                 one.push(1);
46                 zero.push(1);
47             }
48         }else if(s[i]==‘+‘){
49             while(ope.top()!=‘(‘){
50                 calc();
51             }
52             ope.push(s[i]);
53             if(s[i+1]!=‘(‘){
54                 one.push(1);
55                 zero.push(1);
56             }
57         }else if(s[i]==‘*‘){
58             ope.push(s[i]);
59             if(s[i+1]!=‘(‘){
60                 one.push(1);
61                 zero.push(1);
62             }
63         }else if(s[i]==‘)‘){
64             while(ope.top()!=‘(‘){
65                 calc();
66             }
67             ope.pop();
68         }
69     }
70     printf("%d", zero.top());
71     return 0;
72 }

 

2. 先转换成后缀表达式，再计算

 1 #include <bits/stdc++.h>
 2 # define LL long long
 3 using namespace std;
 4 
 5 const int mod=10007;
 6 stack<char> ope;
 7 stack<int> one;
 8 stack<int> zero;
 9 
10 
11 int main(){
12     string s;
13     int L;
14     cin>>L;
15     cin>>s;
16     string str;
17     //补全需要填数的位置
18     for(int i=0;i<L;++i){
19         if(s[i]==‘(‘) {
20             str+=‘(‘;
21         }else {
22             if(i==0 || (i>0 && s[i-1]!=‘)‘)){
23                 str+=‘n‘;
24             }
25             str+=s[i];
26         }
27     }
28     if(str[str.size()-1]==‘+‘ || str[str.size()-1]==‘*‘){
29         str+=‘n‘;
30     }
31 
32     //中缀转后缀
33     vector<char> suff;
34     for(int i=0;i<str.size();++i){
35         if(str[i]==‘n‘){
36             suff.push_back(str[i]);
37         }else if(str[i]==‘(‘ || str[i]==‘*‘){
38             ope.push(str[i]);
39         }else if(str[i]==‘+‘){
40             while(!ope.empty() && ope.top()==‘*‘){
41                 ope.pop();
42                 suff.push_back(‘*‘);
43             }
44             ope.push(str[i]);
45         }else{
46             while(!ope.empty() && ope.top()!=‘(‘){
47                 suff.push_back(ope.top());
48                 ope.pop();
49             }
50             ope.pop();
51         }
52     }
53     while(!ope.empty()){
54         suff.push_back(ope.top());
55         ope.pop();
56     }
57 
58     for(int i=0;i<suff.size();++i){
59         if(suff[i]==‘n‘){
60             one.push(1);
61             zero.push(1);
62         }else if(suff[i]==‘+‘){
63             int o1=one.top(); one.pop();
64             int o2=one.top(); one.pop();
65             int z1=zero.top(); zero.pop();
66             int z2=zero.top(); zero.pop();
67             int t1=((o1*z2)%mod+(z1*o2)%mod+(o1*o2)%mod)%mod;
68             int t0=(z1*z2)%mod;
69             one.push(t1);
70             zero.push(t0);
71         }else if(suff[i]==‘*‘){
72             int o1=one.top(); one.pop();
73             int o2=one.top(); one.pop();
74             int z1=zero.top(); zero.pop();
75             int z2=zero.top(); zero.pop();
76             int t1=(o1*o2)%mod;
77             int t0=((o1*z2)%mod+(z1*o2)%mod+(z1*z2)%mod)%mod;
78             one.push(t1);
79             zero.push(t0);
80         }
81     }
82     printf("%d", zero.top());
83     return 0;
84 }