P1310 表达式的值 （表达式计算）

题目链接

 

解法：

栈

#include <bits/stdc++.h>
# define LL long long
using namespace std;

const int mod=10007;
stack<char> ope;
stack<int> one;
stack<int> zero;

void calc(){
    char o=ope.top();
    ope.pop();
    int one1=one.top();
    one.pop();
    int one2=one.top();
    one.pop();
    int zero1=zero.top();
    zero.pop();
    int zero2=zero.top();
    zero.pop();
    int t1;
    int t0;
    if(o=='*'){
        t1=(one1*one2)%mod;
        t0=((one1*zero2)%mod+(one2*zero1)%mod+(zero1*zero2)%mod)%mod;
    }else {
        t1=((zero1*one2)%mod+(one1*zero2)%mod+(one2*one1)%mod)%mod;
        t0=(zero1*zero2)%mod;
    }
    one.push(t1);
    zero.push(t0);
}

int main(){
    int L;
    cin>>L;
    string s;
    cin>>s;
    s='('+s+')';

    for(int i=0;i<s.size();++i){
        if(s[i]=='('){
            ope.push(s[i]);
            if(s[i+1]!='('){
                one.push(1);
                zero.push(1);
            }
        }else if(s[i]=='+'){
            while(ope.top()!='('){
                calc();
            }
            ope.push(s[i]);
            if(s[i+1]!='('){
                one.push(1);
                zero.push(1);
            }
        }else if(s[i]=='*'){
            ope.push(s[i]);
            if(s[i+1]!='('){
                one.push(1);
                zero.push(1);
            }
        }else if(s[i]==')'){
            while(ope.top()!='('){
                calc();
            }
            ope.pop();
        }
    }
    printf("%d", zero.top());
    return 0;
}

2. 先转换成后缀表达式，再计算

#include <bits/stdc++.h>
# define LL long long
using namespace std;

const int mod=10007;
stack<char> ope;
stack<int> one;
stack<int> zero;


int main(){
    string s;
    int L;
    cin>>L;
    cin>>s;
    string str;
    //补全需要填数的位置
    for(int i=0;i<L;++i){
        if(s[i]=='(') {
            str+='(';
        }else {
            if(i==0 || (i>0 && s[i-1]!=')')){
                str+='n';
            }
            str+=s[i];
        }
    }
    if(str[str.size()-1]=='+' || str[str.size()-1]=='*'){
        str+='n';
    }

    //中缀转后缀
    vector<char> suff;
    for(int i=0;i<str.size();++i){
        if(str[i]=='n'){
            suff.push_back(str[i]);
        }else if(str[i]=='(' || str[i]=='*'){
            ope.push(str[i]);
        }else if(str[i]=='+'){
            while(!ope.empty() && ope.top()=='*'){
                ope.pop();
                suff.push_back('*');
            }
            ope.push(str[i]);
        }else{
            while(!ope.empty() && ope.top()!='('){
                suff.push_back(ope.top());
                ope.pop();
            }
            ope.pop();
        }
    }
    while(!ope.empty()){
        suff.push_back(ope.top());
        ope.pop();
    }

    for(int i=0;i<suff.size();++i){
        if(suff[i]=='n'){
            one.push(1);
            zero.push(1);
        }else if(suff[i]=='+'){
            int o1=one.top(); one.pop();
            int o2=one.top(); one.pop();
            int z1=zero.top(); zero.pop();
            int z2=zero.top(); zero.pop();
            int t1=((o1*z2)%mod+(z1*o2)%mod+(o1*o2)%mod)%mod;
            int t0=(z1*z2)%mod;
            one.push(t1);
            zero.push(t0);
        }else if(suff[i]=='*'){
            int o1=one.top(); one.pop();
            int o2=one.top(); one.pop();
            int z1=zero.top(); zero.pop();
            int z2=zero.top(); zero.pop();
            int t1=(o1*o2)%mod;
            int t0=((o1*z2)%mod+(z1*o2)%mod+(z1*z2)%mod)%mod;
            one.push(t1);
            zero.push(t0);
        }
    }
    printf("%d", zero.top());
    return 0;
}