题目链接

BZOJ2217

题解

如果只判定存不存在方案的话，我倒是想到可以将\(2\)拆成两个\(1\)，其中一个不能作为区间开头，线段树优化计算补集方案数

但是一看这道题要输出方案啊，，，
怎么办？
考虑如果凑不出\(x\)，那一定可以凑出\(x + 1\)
我们就找到前缀和为\(x\)的位置，如果没有，就找\(x + 1\)
前缀和为\(x\)当然就得到答案啦
前缀和为\(x + 1\)，我们考虑将区间整体右移，如果左端点出去和右端点进来的数相同，区间值不变，如果不同，那我们就可以通过调整使得区间的值减少\(1\)
贪心预处理一下答案即可

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
#define lbt(x) (x & -x)
using namespace std;
const int maxn = 1000005,maxm = 100005,INF = 1000000000;
inline int read(){int out = 0,flag = 1; char c = getchar();while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}return out * flag;
}
int ansl[maxn << 1],ansr[maxn << 1],sum[maxn],R[maxn],n,m,N;
char S[maxn];
int main(){n = read(); m = read();scanf("%s",S + 1);REP(i,n) sum[i] = sum[i - 1] + (S[i] == 'W' ? 1 : 2);for (int i = n - 1; ~i; i--){if (sum[i + 1] - sum[i] == 2) R[i] = R[i + 1] + 1;else R[i] = 0;}//REP(i,n + 1) printf("R[%d] = %d\n",i - 1,R[i - 1]);int pos = 1;for (int i = 1; i <= sum[n]; i++){while (sum[pos] != i && sum[pos] != i + 1) pos++;if (sum[pos] == i) ansl[i] = 1,ansr[i] = pos;else {if (R[0] == R[pos]){ansl[i] = R[0] + 2,ansr[i] = pos + R[pos];}else if (R[0] < R[pos]) ansl[i] = R[0] + 2,ansr[i] = pos + R[0];else if (pos + R[pos] < n) ansl[i] = R[pos] + 2,ansr[i] = pos + R[pos] + 1;}}int len;while (m--){len = read();if (!ansl[len] || ansl[len] > ansr[len]) puts("NIE");else printf("%d %d\n",ansl[len],ansr[len]);}return 0;
}