HDU 4891 The Great Pan (题意题+模拟)

题意:给定一个文章,问你有多少种读法,计算方法有两种,如果在$中,如果有多个空格就算n+1,如果是一个就算2的次方,如果在{}中,

那么就是把每个空格数乘起来。

析:直接模拟,每次计算一行,注意上一行最后有空格,下面第一个也是,要全部算上。

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 = 1e3 + 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[1000000];
const int ttt = 100000;
int main(){
    while(scanf("%d", &n) == 1){
        getchar();
        LL ans = 1;
        bool ok = false;
        bool is = false;
        int cnt1 = 0, cnt2 = 0;
        bool ok1 = false;
        bool ok2 = false;
        while(n--){
            gets(s);
            for(int i = 0; s[i]; ++i){
                if(s[i] == '$' && !ok1) ok1 = true;
                else if(s[i] == '{') ok2 = true;
                else if(s[i] == '$'){
                    ok1 = false;
                    if(!cnt1)  continue;
                    if(cnt1 == 1){
                        if(ans * 2LL > ttt) ok = true;
                        else ans *= 2LL;
                    }
                    else {
                        if(ans * (cnt1+1) > ttt)  ok = true;
                        else  ans *= (cnt1 + 1);
                    }
                    cnt1 = 0;
                }
                else if(s[i] == '}'){
                    ok2 = false;
                    if(ans * (cnt2+1) > ttt) ok = true;
                    else ans *= cnt2 + 1;
                    cnt2 = 0;
                }
                else if(s[i] == '|' && ok2)  ++cnt2;
                else if(s[i] == ' ' && ok1)  ++cnt1;
                else{
                    if(ok1 && cnt1){
                        if(cnt1 == 1){
                            if(ans * 2LL > ttt) ok = true;
                            else ans *= 2LL;
                        }
                        else {
                            if(ans * (cnt1+1) > ttt)  ok = true;
                            else  ans *= (cnt1 + 1);
                        }
                        cnt1 = 0;
                    }
                }
            }
        }
        if(ok)  puts("doge");
        else printf("%I64d
", ans);
    }
    return 0;
}
原文地址:https://www.cnblogs.com/dwtfukgv/p/6014124.html