UVaLive 6833 Miscalculation (表达式计算)

题意：给定一个表达式，只有+*，然后问你按照法则运算和从左到右计算结果有什么不同。

析：没什么可说的，直接算两次就好。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#define debug puts("+++++")
//#include <tr1/unordered_map>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
//using namespace std :: tr1;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 2e5 + 5;
const LL mod = 1e9 + 7;
const int N = 1e6 + 5;
const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
inline LL gcd(LL a, LL b){  return b == 0 ? a : gcd(b, a%b); }
inline int gcd(int a, int b){  return b == 0 ? a : gcd(b, a%b); }
inline int lcm(int a, int b){  return a * b / gcd(a, b); }
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
char s[maxn];
stack<char> mark;
stack<int> num;
int ans1, ans2;

void solve(){
    int n = strlen(s);
    s[n] = '+';
    s[n+1] = '0';
    s[n+2] = 0;
    n += 2;
    int cnt = 0;
    ans1 = 0;
    vector<int> v1;
    vector<char> v2;
    for(int i = 0; i < n; ++i){
        if(isalnum(s[i]))  cnt = 10*cnt + s[i] - '0';
        else{
            v1.push_back(cnt);
            v2.push_back(s[i]);
            if(mark.empty()){
                num.push(cnt);
                mark.push(s[i]);
            }
            else{
                char op = mark.top();
                if(op == '*'){
                    int t = num.top();  num.pop();
                    t *= cnt;
                    num.push(t);
                    mark.pop();
                }
                else  num.push(cnt);
                mark.push(s[i]);
            }
            cnt = 0;
        }
    }
    while(!num.empty())  ans1 += num.top(), num.pop();
    ans2 = v1[0];
    cnt = 0;
    for(int i = 1; i < v1.size(); ++i)
        if(v2[cnt++] == '*')  ans2 *= v1[i];
        else ans2 += v1[i];
}

int main(){
    while(scanf("%s", s) == 1){
        scanf("%d", &n);
        while(!mark.empty())  mark.pop();
        solve();
        if(ans1 == ans2 && ans1 == n)  puts("U");
        else if(ans1 != n && ans2 != n)  puts("I");
        else if(ans1 == n)  puts("M");
        else puts("L");
    }
    return 0;
}