[2019杭电多校第七场][hdu6646]A + B = C(hash)

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6646

题意为求$a*10^{x}+b*10^{y}=c*10^{z}$满足公式的任意一组解$x,y,z$。

因为c有可能会由$a+b$进位得到，所以先在c后添加0使得c长度最长，然后先固定a的长度为c-1或c,遍历b的长度为b到c。

用hash判断是否相等。再交换b和a的顺序再判断一次。

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<string>
 #include<algorithm>
 using namespace std;
 typedef long long ll;
 const int maxn = 1e5 + ;
 struct hash {
     int base, mod;
     int tot, Hash[maxn], xp[maxn];
     hash() {
         tot = ; xp[] = ; base = ; mod = 1e9 + ;
     }
     void init() { tot = ; }
     void insert(int c) {
         tot++;
         Hash[tot] = (1LL * Hash[tot - ] * base + c) % mod;
         xp[tot] = (1LL * xp[tot - ] * base) % mod;
     }
     int query(int l, int r) {
         if (l == ) return Hash[r];
         return (Hash[r] - (1LL * Hash[l - ] * xp[r - l + ] % mod) + mod) % mod;
     }
     friend bool check(hash &a, int l1, int r1, hash &b, int l2, int r2, hash &c, int l3, int r3) {
         if ((a.query(l1, r1) + b.query(l2, r2)) % a.mod != c.query(l3, r3)) return false;
         return true;
     }

 }h[];
 char a[][maxn];
 int len[], oldlen[];
 int main() {
     int t;
     scanf("%d", &t);
     while (t--) {
         scanf("%s%s%s", a[] + , a[] + , a[] + );
         for (int i = ; i <= ; i++)
             oldlen[i] = len[i] = strlen(a[i] + ), h[i].init();
         for (int i = ; i <= ; i++)
             for (int j = ; j <= len[i]; j++)
                 h[i].insert(a[i][j] - '');
         int x = , y = , z = ;
         while (len[] <= max(len[], len[]) + ) {
             z++;
             a[][++len[]] = '';
             h[].insert();
         }
         while (len[] < len[]) {
             a[][++len[]] = '';
             h[].insert();
         }
         while (len[] < len[]) {
             a[][++len[]] = '';
             h[].insert();
         }
         int f = ;
         for (int i = oldlen[]; i <= len[] && f; i++) {
             if (check(h[], , i, h[], , len[] - , h[], , len[])) {
                 f = ;
                 x = i - oldlen[], y = len[] -  - oldlen[];
                 break;
             }
             if (check(h[], , i, h[], , len[], h[], , len[])) {
                 f = ;
                 x = i - oldlen[], y = len[] - oldlen[];
                 break;
             }
         }
         for (int i = oldlen[]; i <= len[] && f; i++) {
             if (check(h[], , len[] - , h[], , i, h[], , len[])) {
                 f = ;
                 x = len[] -  - oldlen[], y = i - oldlen[];
                 break;
             }
             if (check(h[], , len[], h[], , i, h[], , len[])) {
                 f = ;
                 x = len[] - oldlen[], y = i - oldlen[];
                 break;
             }
         }
         if (f)
             printf("-1\n");
         else
             printf("%d %d %d\n", x, y, z);
     }
 }