題目鏈接
本題我們可以考慮成從後往前的匹配，因爲要讓後綴最多的爲0，不妨就是從後往前按照需求的進行匹配即可，這個很好想，就是細節有點多了。
先是講講後綴數組的做法，這裏不能用倍增的後綴數組，因爲還要帶一個log，這樣要是寫的不好可就TLE了，所以這裏用線性複雜度的DC3來寫這道題，然後求出height，找到需求串的位置。然後按照sa函數的排名從前往後和從後往前的分別去考慮，進行相同前綴的計算，然後算出對應的前綴的長度和最多允許的長度，然後再看看如果一邊是滿了的時候，可不可以9或者0的進位（如果前面沒有進位的時候，就是找0）。
#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <bitset>
//#include <unordered_map>
//#include <unordered_set>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
#define MP(a, b) make_pair(a, b)
using namespace std;
typedef unsigned long long ull;
typedef unsigned int uit;
typedef long long ll;
const int maxN = 6e6 + 10;
struct DC3
{
static const int maxn = 6e6 + 10;
#define F(x) ((x)/3+((x)%3==1?0:tb))
#define G(x) ((x)<tb?(x)*3+1:((x)-tb)*3+2)
int sa[maxN], rk[maxN], height[maxN], s[maxN];
int wa[maxn], wb[maxn], wv[maxn], wss[maxn];
inline int c0(int *r, int a, int b) { return r[a]==r[b] && r[a+1]==r[b+1] && r[a+2]==r[b+2]; }
inline int c12(int k, int *r, int a, int b)
{
if(k==2)
return r[a]<r[b] || (r[a]==r[b] && c12(1,r,a+1,b+1));
else return r[a]<r[b] || (r[a]==r[b] && wv[a+1] < wv[b+1]);
}
inline void ssort(int *r, int *a, int *b, int n, int m)
{
int i;
for(i=0; i<n; i++) wv[i] = r[a[i]];
for(i=0; i<m; i++) wss[i] = 0;
for(i=0; i<n; i++) wss[wv[i]]++;
for(i=1; i<m; i++) wss[i] += wss[i-1];
for(i=n-1; i>=0; i--)
b[--wss[wv[i]]] = a[i];
}
inline void dc3(int *r, int *sa, int n, int m)
{
int i, j, *rn = r + n;
int *san = sa + n, ta = 0, tb = (n + 1) / 3, tbc = 0, p;
r[n] = r[n + 1] = 0;
for(i=0;i<n;i++) if(i % 3 != 0) wa[tbc++] = i;
ssort(r+2, wa, wb, tbc, m);
ssort(r+1, wb, wa, tbc, m);
ssort(r, wa, wb, tbc, m);
for(p=1, rn[F(wb[0])] = 0, i = 1; i < tbc; i++)
rn[F(wb[i])] = c0(r, wb[i-1], wb[i]) ? p - 1 : p++;
if(p < tbc) dc3(rn, san, tbc, p);
else for(i=0; i<tbc; i++) san[rn[i]] = i;
for(i=0; i<tbc; i++) if(san[i]<tb) wb[ta++] = san[i] * 3;
if(n % 3 == 1) wb[ta++] = n - 1;
ssort(r, wb, wa, ta, m);
for(i=0; i<tbc; i++) wv[wb[i]=G(san[i])]=i;
for(i=0, j=0, p=0; i<ta && j<tbc; p++)
sa[p]=c12(wb[j]%3, r, wa[i], wb[j]) ? wa[i++] : wb[j++];
for(; i<ta; p++) sa[p] = wa[i++];
for(; j<tbc; p++) sa[p] = wb[j++];
}
inline void da(int n,int m)
{
for(int i=n; i<n*3; i++) s[i]=0;
dc3(s, sa, n+1, m);
int i, j, k=0;
for(i=0;i<=n;i++) rk[sa[i]]=i;
for(i=0;i<n;i++){
if(k)k--;
j=sa[rk[i]-1];
while(s[i+k] == s[j+k])k++;
height[rk[i]]=k;
}
}
} sa;
char s[maxN];
int N, a[maxN], b[maxN], ans, num0[maxN], num9[maxN];
bool zp[maxN];
int main()
{
while(scanf("%s", s + 1) != EOF)
{
int len = (int)strlen(s + 1);
for(int i=1; i<=len; i++) a[i] = s[len - i + 1] - '0';
num0[len + 1] = num9[len + 1] = 0;
for(int i=len; i>=1; i--)
{
if(a[i] == 0) num0[i] = num0[i + 1] + 1;
else num0[i] = 0;
if(a[i] == 9) num9[i] = num9[i + 1] + 1;
else num9[i] = 0;
}
bool zero = true;
b[1] = 10 - a[1];
if(b[1] == 10) b[1] = 0;
else zero = false;
zp[1] = zero;
for(int i=2; i<=len; i++)
{
if(zero)
{
b[i] = 10 - a[i];
if(b[i] == 10) b[i] = 0;
else zero = false;
}
else b[i] = 9 - a[i];
zp[i] = zero;
}
sa.s[0] = 12;
for(int i=1; i<=len; i++)
{
sa.s[i] = a[i] + 1;
}
for(int i=1; i<=len; i++)
{
sa.s[len + i] = b[i] + 1;
}
N = len << 1;
sa.s[N + 1] = 12;
sa.s[N + 2] = 12;
sa.da(N + 2, 15);
// for(int i=1; i<=N; i++) printf("%d%c", sa.sa[i], i == N ? '\n' : ' ');
// for(int i=1; i<=N; i++) printf("%d%c", sa.rk[i], i == N ? '\n' : ' ');
// for(int i=1; i<=N; i++) printf("%d%c", sa.height[i], i == N ? '\n' : ' ');
int mid_rk = sa.rk[len + 1];
ans = 0;
for(int i=mid_rk + 1, H = INF, id, id_pre, id_suff, len_pre, len_suff, tmp; i <= N; i++)
{
H = min(H, sa.height[i]);
if(!H) break;
id = sa.sa[i];
if(id > len) continue;
id_pre = H;
id_suff = id;
len_suff = len - id + 1;
len_pre = len - len_suff;
tmp = min(H, min(len_pre, len_suff));
if(tmp == min(len_pre, len_suff))
{
if(len_pre > len_suff)
{
if(zp[len_suff]) tmp += num0[len_suff + 1];
else tmp += min(num9[len_suff + 1], id - tmp - 1);
}
else if(len_suff > len_pre)
{
if(zp[len_pre]) tmp += num0[len_pre + len_pre + 1];
else tmp += num9[len_pre + len_pre + 1];
}
}
ans = max(ans, tmp);
}
for(int i=mid_rk - 1, H = sa.height[mid_rk], id, id_pre, id_suff, len_pre, len_suff, tmp; i >= 1; i--)
{
id = sa.sa[i];
if(id > len) continue;
id_pre = H;
id_suff = id;
len_suff = len - id + 1;
len_pre = len - len_suff;
tmp = min(H, min(len_pre, len_suff));
if(tmp == min(len_pre, len_suff))
{
if(len_pre > len_suff)
{
if(zp[len_suff]) tmp += num0[len_suff + 1];
else tmp += min(num9[len_suff + 1], id - tmp - 1);
}
else if(len_suff > len_pre)
{
if(zp[len_pre]) tmp += num0[len_pre + len_pre + 1];
else tmp += num9[len_pre + len_pre + 1];
}
}
ans = max(ans, tmp);
H = min(H, sa.height[i]);
if(!H) break;
}
printf("%d\n", ans);
}
return 0;
}
/*
2872
ans:2
19919
ans:3
91991
ans:3
9199
ans:1
901
ans:1
9901
ans:2
9911
ans:1
2018
ans:1
109
ans:1
9091
ans:2
12080
ans:2
1000
ans:2
9545455
ans:4
95454545455
ans:6
10090
ans:2
15999841
ans:5
*/
再講講exkmp的做法，如果使用擴展kmp的話，我們可以先去一樣的處理出本串對於查詢串（也就是需求串的）的最大匹配，也就是每個後綴的最長前綴長度，ex數組。然後與後綴數組相同的來去進行判斷。
#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <bitset>
#include <unordered_map>
#include <unordered_set>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f
#define eps 1e-6
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
using namespace std;
typedef unsigned long long ull;
typedef unsigned int uit;
typedef long long ll;
//const int maxN = 100;
const int maxN = 1e6 + 7;
char A[maxN], B[maxN];
bool Zop[maxN];
int sa[maxN], sb[maxN];
int lena, lenb, num0[maxN], num9[maxN];
int p[maxN], ex[maxN];
//p數組是用來讓B串自己匹配自己的
void exkmp()
{
p[1] = lenb;
int x = 1;
while(sb[x] == sb[x + 1] && x + 1 <= lenb) x++; //因爲我們p[1]是具有一定性，所以我們不能直接用，所以要先暴力求出p[2]
p[2] = x - 1;
int k = 2;
for(int i=3; i<=lenb; i++)
{
int pp = k + p[k] - 1, L = p[i - k + 1]; //pp實際上是p
if(i + L < pp + 1) p[i] = L;//i-k+L<pp-k+1化簡後i+L<pp
else
{
int j = pp - i + 1;
if(j < 0) j = 0;
while(sb[j + 1] == sb[i + j] && i + j <= lenb) j++;
p[i] = j;
k = i;
}
}
x = 1;
while(sa[x] == sb[x] && x <= lenb) x++;//ex[1]並不具有一定性，所以我們暴力求出ex[1]
ex[1] = x - 1;
k = 1;
for(int i=2; i<=lena; i++)
{
int pp = k + ex[k] - 1, L = p[i - k + 1];
if(i + L < pp + 1) ex[i] = L;
else
{
int j = pp - i + 1;
if(j < 0) j = 0;
while(sb[j + 1] == sa[i + j] && i + j <= lena && j <= lenb) j++;
ex[i] = j;
k = i;
}
}
}
int main()
{
while(scanf("%s", A + 1) != EOF)
{
lena = (int)strlen(A + 1); lenb = lena;
for(int i=1; i<=lena; i++)
{
sa[i] = A[lena + 1 - i] - '0';
}
for(int i=lena, pos0 = lena, pos9 = lenb; i>=1; i--)
{
if(sa[i] == 0)
{
pos9 = i - 1;
}
else if(sa[i] == 9)
{
pos0 = i - 1;
}
else { pos0 = pos9 = i - 1; }
num0[i] = pos0 - i + 1;
num9[i] = pos9 - i + 1;
}
bool zero = true;
sb[1] = 10 - sa[1];
if(sb[1] == 10) sb[1] = 0;
else zero = false;
for(int i=2; i<=lenb; i++)
{
if(zero)
{
sb[i] = 10 - sa[i];
if(sb[i] == 10) sb[i] = 0;
else zero = false;
}
else
{
sb[i] = 9 - sa[i];
}
Zop[i] = zero;
}
exkmp();
// printf("bp: "); for(int i=1; i<=lenb; i++) printf("%d%c", p[i], i == lenb ? '\n' : ' ');
// printf("ex: "); for(int i=1; i<=lena; i++) printf("%d%c", ex[i], i == lena ? '\n' : ' ');
int ans = 0;
for(int i=2, len_pre, len_suff; i<=lena; i++)
{
len_pre = ex[i];
len_pre = min(len_pre, i - 1);
len_suff = lena - i + 1;
if(len_pre < i - 1)
{
if(len_pre >= len_suff)
{
ans = max(ans, len_pre + min(num9[len_pre + 1], i - 1 - len_pre));
}
else
{
ans = max(ans, len_pre);
}
}
else
{
if(i + len_pre == lena + 1) ans = max(ans, len_pre);
else if(Zop[i + len_pre - 1])
{
ans = max(ans, len_pre + num0[i + len_pre]);
}
else
{
ans = max(ans, len_pre + num9[i + len_pre]);
}
}
}
printf("%d\n", ans);
}
return 0;
}
/*
15999841
ans:5
9901
ans:2
95454545455
ans:6
*/