[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 + 110;
struct hash {
    int base, mod;
    int tot, Hash[maxn], xp[maxn];
    hash() {
        tot = 0; xp[0] = 1; base = 10; mod = 1e9 + 7;
    }
    void init() { tot = 0; }
    void insert(int c) {
        tot++;
        Hash[tot] = (1LL * Hash[tot - 1] * base + c) % mod;
        xp[tot] = (1LL * xp[tot - 1] * base) % mod;
    }
    int query(int l, int r) {
        if (l == 1) return Hash[r];
        return (Hash[r] - (1LL * Hash[l - 1] * xp[r - l + 1] % 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[4];
char a[4][maxn];
int len[4], oldlen[4];
int main() {
    int t;
    scanf("%d", &t);
    while (t--) {
        scanf("%s%s%s", a[1] + 1, a[2] + 1, a[3] + 1);
        for (int i = 1; i <= 3; i++)
            oldlen[i] = len[i] = strlen(a[i] + 1), h[i].init();
        for (int i = 1; i <= 3; i++)
            for (int j = 1; j <= len[i]; j++)
                h[i].insert(a[i][j] - '0');
        int x = 0, y = 0, z = 0;
        while (len[3] <= max(len[1], len[2]) + 1) {
            z++;
            a[3][++len[3]] = '0';
            h[3].insert(0);
        }
        while (len[1] < len[3]) {
            a[1][++len[1]] = '0';
            h[1].insert(0);
        }
        while (len[2] < len[3]) {
            a[2][++len[2]] = '0';
            h[2].insert(0);
        }
        int f = 1;
        for (int i = oldlen[1]; i <= len[1] && f; i++) {
            if (check(h[1], 1, i, h[2], 1, len[2] - 1, h[3], 1, len[3])) {
                f = 0;
                x = i - oldlen[1], y = len[2] - 1 - oldlen[2];
                break;
            }
            if (check(h[1], 1, i, h[2], 1, len[2], h[3], 1, len[3])) {
                f = 0;
                x = i - oldlen[1], y = len[2] - oldlen[2];
                break;
            }
        }
        for (int i = oldlen[2]; i <= len[2] && f; i++) {
            if (check(h[1], 1, len[1] - 1, h[2], 1, i, h[3], 1, len[3])) {
                f = 0;
                x = len[1] - 1 - oldlen[1], y = i - oldlen[2];
                break;
            }
            if (check(h[1], 1, len[1], h[2], 1, i, h[3], 1, len[3])) {
                f = 0;
                x = len[1] - oldlen[1], y = i - oldlen[2];
                break;
            }
        }
        if (f)
            printf("-1\n");
        else
            printf("%d %d %d\n", x, y, z);
    }
}