【CSP】90%解压缩&80%化学方程式 两道大模拟
大模拟就没什么太多说的了，《解压缩》需要用一点位运算，《化学方程式》需要用字符串。
学到的主要经验就是，使用次数>2的功能拆分到不同函数内，并且对于难写的函数提前单独调试错误，不要等写完了再debug。
对于《化学方程式》，我使用模块化思想。第一步拆成不同的项，把系数缓存起来，然后对项（如Au(Ag(CN)2) ）先序列化（S = Au,"(",Ag,"(",C,N,")",2,")"），然后对其进行括号栈分析。
#include <cstdio> #include <iostream> #include <algorithm> #include <vector> #include <set> #include <string.h> #define up(l,r,i) for(int i=l;i<=r;i++) #define dn(l,r,i) for(int i=r;i>=l;i--) typedef long long ll; using namespace std; inline int _max(const int& a,const int& b){return a>b?a:b;} inline int _min(const int& a,const int& b){return a<b?a:b;} const int MAXN = 3000020; inline int normal(int& x){ x = x&256; return x; } int p2[12] = {0,1,3,7,15,31,63,127,255,511,1023,2047}; int bts[MAXN],s; int op[MAXN],ocnt; int total = 0; inline int highn(int x,int n){ return x>>(8-n); } inline int lown(int x,int n){ return x&p2[n]; } void prbyte(int x){ cout<<"["; bool tmp[8]; dn(0,7,i){ tmp[i] = x%2; x/=2; } up(0,7,i) cout<<tmp[i]; cout<<"]"; } int pcnt = 0; void pbyte16(int x){ if(pcnt >= 8 && ocnt < total-1){ cout<<endl; pcnt = 0; } int a = highn(x,4); int b = lown(x,4); char c,d; if(a > 9) c = a-10+'a'; else c = a+'0'; if(b > 9) d = b-10+'a'; else d = b+'0'; cout<<c<<d; pcnt++; } inline int getbyte(char a,char b){ int ret = 0; if(a >= '0' && a <= '9') ret += (int)(a-'0')*16; else ret += (int)(10+a-'a')*16; if(b >= '0' && b <= '9') ret += (int)(b-'0'); else ret += (int)(10+b-'a'); return ret; } void element(int &cnt){ int l = 0; if(lown(bts[cnt],2) == 0){ l = highn(bts[cnt],6); //printf("\ncnt=%d,进入字面值,高六位为%d\n",cnt,l); int del = l - 59; int k = 1; cnt++; if(del > 0) l = 0; while(del>0){ l += k*bts[cnt]; k *= 256; del--; cnt++; } l++; //printf("\ncnt=%d,字面值长度l=%d\n",cnt,l); while(l){ l--; //printf("[l=%d]",l); op[ocnt++] = bts[cnt]; pbyte16(bts[cnt]); cnt++; } } else if(lown(bts[cnt],2) == 1){ int o; o = bts[cnt+1]; o += highn(bts[cnt],3)*256; l = lown(bts[cnt],5)>>2; l += 4; //printf("\ncnt=%d,进入回溯引用1,l=%d,o=%d\n",cnt,l,o); int cnt2 = ocnt-o;//-1+1 int tmp = ocnt-1; while(l){ if(cnt2 > tmp) cnt2 = tmp-o+1; op[ocnt++] = op[cnt2++]; pbyte16(op[ocnt-1]); l--; } cnt += 2; } else if(lown(bts[cnt],2) == 2){ l = highn(bts[cnt],6)+1; int o = bts[++cnt]; o += bts[++cnt]*256; int cnt2 = ocnt-o;//-1+1 int tmp = ocnt-1; //printf("\ncnt=%d,进入回溯引用2,l=%d,o=%d\n",cnt,l,o); while(l){ if(cnt2 > tmp) cnt2 = tmp-o+1; op[ocnt++] = op[cnt2++]; pbyte16(op[ocnt-1]); l--; } cnt++; } } void solve(){ int cnt = 0; int k = 1; while(highn(bts[cnt],1)){ total += k*lown(bts[cnt],7); cnt++; k *= 128; } total += k*lown(bts[cnt],7); //printf("\ncnt=%d,引导区结束\n",cnt); cnt++;//现在cnt为数据域的第一位 while(cnt < s-1){ element(cnt); } } int main() { //freopen("y.in","r",stdin); //freopen("y.out","w",stdout); //ios::sync_with_stdio(false); cin>>s; getchar(); char a,b; up(0,s-1,i){ a = getchar(); if(a == '\n') {i--;continue;} b = getchar(); bts[i] = getbyte(a,b); } solve(); return 0; }
#include <cstdio> #include <iostream> #include <algorithm> #include <vector> #include <set> #include <string.h> #include <map> #include <stdlib.h> #include <vector> #define up(l,r,i) for(int i=l;i<=r;i++) #define dn(l,r,i) for(int i=r;i>=l;i--) typedef long long ll; using namespace std; inline int _max(const int& a,const int& b){return a>b?a:b;} inline int _min(const int& a,const int& b){return a<b?a:b;} int state = 0;//0 - > 等待新项 //1 - > 正在读取项 int N; inline bool isnum(char c){ return (c>='0'&&c<='9'); } map<string,int> equa; struct Ele{ int type;//0 == num, 1 = element, -n~n为括号级数(n>=2) int n; string ele; }; struct Pr{ int f,s; }; Ele create(int t,int n,string e){ Ele ret; ret.type = t; ret.n = n; ret.ele = e; return ret; } bool solve(string &s){ int len = s.length(); int k = 1; int ca = 0;//缓存 up(0,len-1,i){ if(state == 0){ if(isnum(s[i])){ int off = 1; while(i+off<len && isnum(s[i+off])){ off++; } string num = s.substr(i,off); ca = atoi(num.c_str()); i += off; i--;//因为i++ state = 1; } else{ state = 1; ca = 1; i--; } } else if(state == 1){ int off = 1; while(i+off<len && s[i+off]!='+' && s[i+off]!='='){ off++; } string ele = s.substr(i,off); //TODO = 实现计算元素 vector<Ele> seq; vector<int> sta; int cnt = 2; string x = ""; up(0,ele.length()-1,j){ if(ele[j] == '(') { seq.push_back(create(cnt,0,x)); sta.push_back(cnt++); } else if (ele[j] == ')'){ int c = sta[sta.size()-1]; sta.pop_back(); seq.push_back(create(-c,0,x)); } else if(isnum(ele[j])){ int of = 1; while(j+of<len && isnum(ele[j+of])){ of++; } string num = ele.substr(j,of); int numnum = atoi(num.c_str()); j += of; j--;//因为i++ seq.push_back(create(0,numnum,x)); } else{ string ee; ee += ele[j]; if(j+1<ele.length() && ele[j+1] >= 'a' && ele[j+1] <= 'z') {ee += ele[j+1];j++;} seq.push_back(create(1, 1, ee)); } } for(auto it = seq.begin(); it != seq.end() ; it++){ if(it->type == 1) continue; else if(it->type == 0){ auto it2 = it-1; if(it2->type == 1){ it2->n *= it->n; } } else{ if(it == seq.end()-1) continue; if((it+1)->type == 1) continue; int kk = (it+1)->n; int cnt = it->type; for(auto it2 = it-1; it2 != seq.begin()-1; it2--){ if(it2->type == -cnt) break; if(it2->type == 1) it2->n *= kk; } } } /*for(auto it = seq.begin(); it != seq.end() ; it++){ if(it->type == 1){ printf("[%s,%d]",it->ele.c_str(),it->n); } }*/ for(auto it = seq.begin(); it != seq.end() ; it++){ if(it->type == 1){ //printf("k=%d,ca=%d\n",k,ca); equa[it->ele] += k*ca*it->n; } } if(s[i+off] == '='){ k = -1; } state = 0; i += off; } } for(auto it = equa.begin();it != equa.end(); it++){ //printf("[%s,%d]",it->first.c_str(),it->second); if(it->second != 0) {return false;} } return true; } int main() { //freopen("y.in","r",stdin); //ios::sync_with_stdio(false); cin>>N; getchar(); up(1,N,i){ string s; getline(cin,s); cout<<(solve(s)?"Y":"N")<<endl; equa.clear(); } return 0; }
虽说不太愿意写， 但听歌写这种玩意反而有种奇妙的快感（？
P.S.以前不重视STL，现在发现真好用。另外ios::sync_with_stdio(false)在ccf的机器上会有时会导致爆0，惨痛教训。
