CodeForces - 508E_Arthur and Brackets 括號問題(stack)

題目大意：給定括號總數n，再給出每組括號的左右半邊的下標相差範圍L, R，求滿足條件的括號序列，若不存在則輸出IMPOSSIBLE

pos[i]:第i個括號的左半邊的位置
len:當前字符串長度
括號匹配成功的條件：pos[i]+L[i]<=len+1<=pos[i]+R[i];

#include <cstdio>
#include <stack>

using namespace std;
const int N = 610;
char ans[2*N];
int len;
int L[N], R[N], pos[N];

int main()
{
    int n;
    scanf("%d", &n);
    len = 0;
    stack <int> s;
    for(int i = 1; i <= n; ++i)
    {
        scanf("%d %d", &L[i], &R[i]);
        s.push(i);
        ans[++len] = '(';
        pos[i] = len;//棧頂元素的位置
        while(!s.empty() && pos[s.top()]+L[s.top()]<=len+1)
        {
            if(pos[s.top()]+R[s.top()] < len+1)
                return 0*printf("IMPOSSIBLE\n");
            ans[++len] = ')';
            s.pop();
        }
    }
    ans[++len] = '\0';
    if(s.empty()) puts(ans+1);
    else printf("IMPOSSIBLE\n");
    return 0;
}