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CodeForces - 200D
Description Recently, Valery have come across an entirely new programming language. Most of all the language attracted him with template functions and procedures. Let us remind you that templates are tools of a language, designed to encode generic algorithms, without reference to some parameters (e.g., data types, buffer sizes, default values). Valery decided to examine template procedures in this language in more detail. The description of a template procedure consists of the procedure name and the list of its parameter types. The generic type T parameters can be used as parameters of template procedures. A procedure call consists of a procedure name and a list of variable parameters. Let‘s call a procedure suitable for this call if the following conditions are fulfilled:
You are given a description of some set of template procedures. You are also given a list of variables used in the program, as well as direct procedure calls that use the described variables. For each call you need to count the number of procedures that are suitable for this call. Input The first line contains a single integer n (1?≤?n?≤?1000) — the number of template procedures. The next n lines contain the description of the procedures specified in the following format: "void procedureName (type_1, type_2, ..., type_t)" (1?≤?t?≤?5), where void is the keyword, procedureName is the procedure name, type_i is the type of the next parameter. Types of language parameters can be "int", "string", "double", and the keyword "T", which denotes the generic type. The next line contains a single integer m (1?≤?m?≤?1000) — the number of used variables. Next m lines specify the description of the variables in the following format: "type variableName", where type is the type of variable that can take values "int", "string", "double", variableName — the name of the variable. The next line contains a single integer k (1?≤?k?≤?1000) — the number of procedure calls. Next k lines specify the procedure calls in the following format: "procedureName (var_1, var_2, ..., var_t)" (1?≤?t?≤?5), where procedureName is the name of the procedure, var_i is the name of a variable. The lines describing the variables, template procedures and their calls may contain spaces at the beginning of the line and at the end of the line, before and after the brackets and commas. Spaces may be before and after keyword void. The length of each input line does not exceed100 characters. The names of variables and procedures are non-empty strings of lowercase English letters and numbers with lengths of not more than 10 characters. Note that this is the only condition at the names. Only the specified variables are used in procedure calls. The names of the variables are distinct. No two procedures are the same. Two procedures are the same, if they have identical names and identical ordered sets of types of their parameters. Output On each of k lines print a single number, where the i-th number stands for the number of suitable template procedures for the i-th call. Sample Input
Input
4 void f(int,T) void f(T, T) void foo123 ( int, double, string,string ) void p(T,double) 3 int a string s double x123 5 f(a, a) f(s,a ) foo (a,s,s) f ( s ,x123) proc(a)
Output
2 1 0 1 0
Input
6 void f(string,double,int) void f(int) void f ( T ) void procedure(int,double) void f (T, double,int) void f(string, T,T) 4 int a int x string t double val 5 f(t, a, a) f(t,val,a) f(val,a, val) solve300(val, val) f (x)
Output
1 3 0 0 2 Source
这道题目首先告诉大家，要有耐心
现在说一下我解题思路，可能麻烦，如果有大神飘过，求指点
先将void f(int , T)变为voidf(int,T)将所有的空格都去掉，然后将字符串的前面四个去掉因为void不是函数名，在‘(‘左边就是函数名。
接着其他的处理也是类似的模拟
/*
Author: 2486
Memory: 256 KB Time: 62 MS
Language: GNU G++ 4.9.2 Result: Accepted
*/
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <map>
#include <string>
#include <iostream>
using namespace std;
const int maxn=1000+5;
struct func {
char name[100+5];
vector<int>G;
} funcs[maxn];
int n,m;
map<string,int>F;
map<string,bool>KS;
char str[maxn];
void getgs(char *st,int l) {
int cnt=0;
for(int i=0; i<l; i++) {
if(st[i]!=' ') {
st[cnt++]=st[i];
}
}
st[cnt]='\0';
}
void getsg(char *st,int id) {
int l=strlen(st);
int cnt=0;
for(int i=4; i<l; i++) {
if(st[i]=='(')break;
funcs[id].name[cnt++]=st[i];
}
funcs[id].name[cnt]='\0';
for(int i=cnt+4; i<l; i++) {
if(str[i-1]=='n'&&st[i]=='t') {
funcs[id].G.push_back(1);
} else if(str[i-1]=='d'&&st[i]=='o') {
funcs[id].G.push_back(2);
} else if(str[i-1]=='s'&&st[i]=='t') {
funcs[id].G.push_back(3);
} else if(str[i]=='T') {
funcs[id].G.push_back(4);
}
}
}
int getsgs(char *st) {
func s;
int l=strlen(st);
int cnt=0;
for(int i=0; i<l; i++) {
if(st[i]=='(')break;
s.name[cnt++]=st[i];
}
s.name[cnt]='\0';
if(!KS[s.name])return 0;
char op[100+5];
int sf=0;
for(int i=cnt+1; i<l; i++) {
if(st[i]==','||st[i]==')') {
op[sf]='\0';
s.G.push_back(F[op]);
sf=0;
continue;
}
op[sf++]=st[i];
}
int res=0;
bool flag;
for(int i=0; i<n; i++) {
if(strcmp(funcs[i].name,s.name)==0) {
flag=true;
if(s.G.size()!=funcs[i].G.size())continue;
for(int j=0; j<s.G.size(); j++) {
if(funcs[i].G[j]==4)continue;
if(funcs[i].G[j]!=s.G[j]) {
flag=false;
break;
}
}
if(flag)res++;
}
}
return res;
}
int main() {
//freopen("D://imput.txt","r",stdin);
scanf("%d",&n);
for(int i=0; i<n; i++) {
gets(str);
int l=strlen(str);
if(l==0) {
i--;
continue;
}
getgs(str,l);
getsg(str,i);
KS[funcs[i].name]=1;
}
scanf("%d",&m);
char op[10],ops[100+5];
for(int i=0; i<m; i++) {
scanf("%s%s",op,ops);
if(op[0]=='d'&&op[1]=='o') {
F[ops]=2;
} else if(op[0]=='i'&&op[1]=='n'&&op[2]=='t') {
F[ops]=1;
} else if(op[0]=='s'&&op[1]=='t') {
F[ops]=3;
}
}
scanf("%d",&m);
for(int i=0; i<m; i++) {
gets(str);
int l=strlen(str);
if(l==0) {
i--;
continue;
}
getgs(str,l);
printf("%d\n",getsgs(str));
}
return 0;
} |
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CodeForces - 200DProgramming Language纯模拟
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原文地址：http://blog.csdn.net/qq_18661257/article/details/47266935
