對大模擬抱有深深的恐懼……不過這次寫好像還好？拿個棧維護一下循環的嵌套，然後重定義一下讀入即可。記得去年在考場上面死活調不粗來，代碼也奇醜無比……希望今年能好一點吧！

#include <bits/stdc++.h>
using namespace std;
#define maxn 100000
int L, mark[maxn];
char ans[maxn];

int fread(int t)
{
    int x = 0, k = 1;
    char c; c = ans[t]; t ++;
    
 
while(t < strlen(ans) && (c < ‘0‘ || c > ‘9‘)) 
    { 
        if(c == ‘-‘) k = -1; 
        if(c == ‘n‘) x = 100;
        c = ans[t]; t ++; 
    }
    while(t < strlen(ans) && c >= ‘0‘ && c <= ‘9‘) 
        x = x * 10 + c - ‘0‘, c = ans[t], t ++;
    return
 
 x * k;
}

int read()
{
    int x = 0, k = 1;
    char c; c = getchar();
    while(c < ‘0‘ || c > ‘9‘) 
    { 
        if(c == ‘-‘) k = -1; 
        if(c == ‘n‘) { x = 1000; return x; }
        c = getchar(); 
    }
    while(c >= ‘0‘ && c <= ‘9‘) x = x * 10 + c - ‘0‘, c = getchar();
    
 
return x * k;
}

struct node
{
    char c; int num;
    node(char _c = ‘z‘ + 1, int _num = 0) 
    { c = _c, num = _num; }
}S[maxn];

void Work()
{
    memset(mark, 0, sizeof(mark)); 
    int cal = 0, top = 0, cnt = 0, err = 0, flag = 0;
    for(int i = 1; i <= L; i ++)
    {
        char c; cin >> c;
        if(c == ‘F‘)
        {
            char x; cin >> x;
            if(mark[x - ‘a‘]) flag = 1;
            int l = read(), r = read();
            mark[x - ‘a‘] = 1;
            if(r - l > 100 && !err) cnt ++, cal = max(cal, cnt);
            if(l > r) err ++;
            S[++ top] = node(x, r - l);
        }
        else
        {
            if(!top) { flag = 1; continue; }
            mark[S[top].c - ‘a‘] = 0;
            if(!err) cnt -= S[top].num > 100;
            if(S[top].num < 0) err --;
            top --;
        }
    }
    int len = strlen(ans), tag = 0, t = 0;
    for(int i = 0; i < len; i ++) 
        if(ans[i] == ‘n‘) tag = i + 2;
        else if(tag && i == tag) { t = fread(i); break; }
    if(top) flag = 1;
    if(flag) { printf("ERR\n"); return; }
    if(cal != t) flag = 1;
    if(flag) { printf("No\n"); return; }
    printf("Yes\n"); return;
}

int main()
{
    int T; scanf("%d", &T);
    while(T --)
    {
        scanf("%d %s", &L, ans);
        Work();
    }
    return 0;
}

【題解】NOIP2017時間復雜度