Codeforces Round #506 D. Concatenated Multiples题解

一、传送门

http://codeforces.com/contest/1029/problem/D

二、题意

给你$N$个数字$a_1,a_2,\cdots,a_n$，一个$K$，求所有$i \ne j$且$(a[i]*10^{len(a[j])}+a[j])%K=0$的pair的对数。

范围：$1 \le N \le 2*10^5$，$1 \le a[i] \le 10^9$，$1 \le K \le 10^9$。

三、思路

记录所有长度$i$，模$K$为$j$的数字的个数。用map记录，显然，键的种类不超过$2*10^5$。枚举map的所有键值对，假设键为$a$，解方程：$(a*10^x+b)%K=0(%K)$，当$x$已知时，$b=-a*10^x(%K)=(K-a*10^x)(%K)$。因为$1\le a[i] \le 10^9$，所以，$x$不超过$10$，可以枚举。然后已知$x$和$b$以后，作为键去map里面找出相应的值，和之前枚举的键值对的值相乘即可。

最后，因为有条件$i \ne j$，所以，再扫描一遍序列$a$，如果$(a[i]*10^{len(a[i])}+a[i])%K=0$，那么，答案减一。

但是，如果用pair<int, int>做map的键，上述思路的复杂度是$O(N*log(N)*log(N)*maxlen)$。这么大的复杂度我当时也想都没想就写了，我天，想什么呢。然后，我被Hack的TLE了。T_T。

那么，思路可以不变，把复杂度降一下就行了。把map的改为unordered_map，此时，键不能是pair，那就把记录个数的map改为$10$个unordered_map<int,long long>就行了。这样，需要枚举两层长度，再需要枚举长度为$i$下的unordered_map，总的复杂度变成了$O(N*maxlen*maxlen)$。那么，最差的ACM版复杂度就是$(N*10*10)$了，稳得很。

四、坑点

1、无论是map还是unordered_map，在迭代遍历的时候，千万不能通过方括号[]的方式去索引值。否则，如果键不存在，内置STL将会把键插进去，导致迭代次数和实际的size不相同。

2、预处理$10^i$，$i$最大是$10$，不是$9$。

3、最后扫描去掉$i=j$的情况时，因为$1 \le a[i] \le 10^9$，$10^{max\{len(a[i])\}}=10^10$，相乘等于$10^{19}$，大于long long的范围，会溢出，所以，一定要记得先把每一项$% K$，再相乘。否则，会挂在我代码最后面注释掉的那个样例。

五、代码(前面一大段都是没用的输入挂，有效代码在最后面)

#include<bits/stdc++.h>
using namespace std;
#define pb(x) push_back(x)
#define mk(x, y) make_pair(x, y)
#define pln() putchar('\n')
#define cln() (cout << '\n')
#define fi first
#define se second
#define MOD 1000000007LL
typedef long long LL;
typedef pair<int, int> PII;
;

/*********************************************************************/
namespace fastIO {
#define BUF_SIZE 100000
#define OUT_SIZE 100000
#define ll long long
//fread->read
    ;
    inline char nc() {
        static char buf[BUF_SIZE], *p1 = buf + BUF_SIZE, *pend = buf + BUF_SIZE;
        if (p1 == pend) {
            p1 = buf; pend = buf + fread(buf, , BUF_SIZE, stdin);
            ; ;}
            //{printf("IO error!\n");system("pause");for (;;);exit(0);}
        }
        return *p1++;
    }
    inline bool blank(char ch) {return ch == ' ' || ch == '\n' || ch == '\r' || ch == '\t';}
    inline void read(int &x) {
        ; ;
        for (; blank(ch); ch = nc());
        if (IOerror)return;
        , ch = nc();
         + ch - ';
        if (sign)x = -x;
    }
    inline void read(ll &x) {
        ; ;
        for (; blank(ch); ch = nc());
        if (IOerror)return;
        , ch = nc();
         + ch - ';
        if (sign)x = -x;
    }
    inline void read(double &x) {
        ; ;
        for (; blank(ch); ch = nc());
        if (IOerror)return;
        , ch = nc();
         + ch - ';
        if (ch == '.') {
            ; ch = nc();
            ');
        }
        if (sign)x = -x;
    }
    inline void read(char *s) {
        char ch = nc();
        for (; blank(ch); ch = nc());
        if (IOerror)return;
        for (; !blank(ch) && !IOerror; ch = nc()) * s++ = ch;
        *s = ;
    }
    inline void read(char &c) {
        for (c = nc(); blank(c); c = nc());
        ; return;}
    }
//getchar->read
    inline void read1(int &x) {
        ; x = ;
        ;
         + ch - ', ch = getchar());
        if (bo)x = -x;
    }
    inline void read1(ll &x) {
        ; x = ;
        ;
         + ch - ', ch = getchar());
        if (bo)x = -x;
    }
    inline void read1(double &x) {
        ; x = ;
        ;
         + ch - ', ch = getchar());
        if (ch == '.') {
            ;
            '), ch = getchar());
        }
        if (bo)x = -x;
    }
    inline void read1(char *s) {
        char ch = getchar();
        for (; blank(ch); ch = getchar());
        for (; !blank(ch); ch = getchar()) * s++ = ch;
        *s = ;
    }
    inline void read1(char &c) {for (c = getchar(); blank(c); c = getchar());}
//scanf->read
    inline void read2(int &x) {scanf("%d", &x);}
    inline void read2(ll &x) {
#ifdef _WIN32
        scanf("%I64d", &x);
#else
#ifdef __linux
        scanf("%lld", &x);
#else
        puts("error:can't recognize the system!");
#endif
#endif
    }
    inline void read2(double &x) {scanf("%lf", &x);}
    inline void read2(char *s) {scanf("%s", s);}
    inline void read2(char &c) {scanf(" %c", &c);}
    inline void readln2(char *s) {gets(s);}
//fwrite->write
    struct Ostream_fwrite {
        char *buf, *p1, *pend;
        Ostream_fwrite() {buf = new char[BUF_SIZE]; p1 = buf; pend = buf + BUF_SIZE;}
        void out(char ch) {
            if (p1 == pend) {
                fwrite(buf, , BUF_SIZE, stdout); p1 = buf;
            }
            *p1++ = ch;
        }
        void print(int x) {
            ], *s1; s1 = s;
            ';
            )out('-'), x = -x;
             + ;
            while(s1-- != s)out(*s1);
        }
        void println(int x) {
            ], *s1; s1 = s;
            ';
            )out('-'), x = -x;
             + ;
            while(s1-- != s)out(*s1); out('\n');
        }
        void print(ll x) {
            ], *s1; s1 = s;
            ';
            )out('-'), x = -x;
             + ;
            while(s1-- != s)out(*s1);
        }
        void println(ll x) {
            ], *s1; s1 = s;
            ';
            )out('-'), x = -x;
             + ;
            while(s1-- != s)out(*s1); out('\n');
        }
        void print(double x, int y) {
            , , , , , , , , ,
                               , 10000000000LL, 100000000000LL, 1000000000000LL, 10000000000000LL,
                               100000000000000LL, 1000000000000000LL, 10000000000000000LL, 100000000000000000LL
                              };
            )out('-'), x = -x; x *= mul[y];
            ll x1 = (ll)floor(x);
            if (x - floor(x) >= 0.5)++x1;
            ll x2 = x1 / mul[y], x3 = x1 - x2 * mul[y]; print(x2);
            ) {; i < y && x3 * mul[i] < mul[y]; '), ++i); print(x3);}
        }
        void println(double x, int y) {print(x, y); out('\n');}
        void print(char *s) {while (*s)out(*s++);}
        void println(char *s) {while (*s)out(*s++); out('\n');}
        , p1 - buf, stdout); p1 = buf;}}
        ~Ostream_fwrite() {flush();}
    } Ostream;
    inline void print(int x) {Ostream.print(x);}
    inline void println(int x) {Ostream.println(x);}
    inline void print(char x) {Ostream.out(x);}
    inline void println(char x) {Ostream.out(x); Ostream.out('\n');}
    inline void print(ll x) {Ostream.print(x);}
    inline void println(ll x) {Ostream.println(x);}
    inline void print(double x, int y) {Ostream.print(x, y);}
    inline void println(double x, int y) {Ostream.println(x, y);}
    inline void print(char *s) {Ostream.print(s);}
    inline void println(char *s) {Ostream.println(s);}
    inline void println() {Ostream.out('\n');}
    inline void flush() {Ostream.flush();}
//puts->write
    char Out[OUT_SIZE], *o = Out;
    inline void print1(int x) {
        ];
        char *p1 = buf;
        ';
        )*o++ = '-', x = -x;
         + ;
        while(p1-- != buf)*o++ = *p1;
    }
    inline void println1(int x) {print1(x); *o++ = '\n';}
    inline void print1(ll x) {
        ];
        char *p1 = buf;
        ';
        )*o++ = '-', x = -x;
         + ;
        while(p1-- != buf)*o++ = *p1;
    }
    inline void println1(ll x) {print1(x); *o++ = '\n';}
    inline void print1(char c) {*o++ = c;}
    inline void println1(char c) {*o++ = c; *o++ = '\n';}
    inline void print1(char *s) {while (*s)*o++ = *s++;}
    inline void println1(char *s) {print1(s); *o++ = '\n';}
    inline void println1() {*o++ = '\n';}
    inline ) == ; puts(Out);}}
    struct puts_write {
        ~puts_write() {flush1();}
    } _puts;
    inline void print2(int x) {printf("%d", x);}
    inline void println2(int x) {printf("%d\n", x);}
    inline void print2(char x) {printf("%c", x);}
    inline void println2(char x) {printf("%c\n", x);}
    inline void print2(ll x) {
#ifdef _WIN32
        printf("%I64d", x);
#else
#ifdef __linux
        printf("%lld", x);
#else
        puts("error:can't recognize the system!");
#endif
#endif
    }
    inline void println2(ll x) {print2(x); printf("\n");}
    inline void println2() {printf("\n");}
#undef ll
#undef OUT_SIZE
#undef BUF_SIZE
};
namespace slowIO {
    template <class T> inline void read(T &x) {
        int t;
        bool flag = false;
        ')) ;
        ';
         + t - ';
        if(flag) x = -x;
    }
};
/*********************************************************************/
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using namespace fastIO;
//using namespace slowIO;
LL n, k, a[MAXN], p10[], len[MAXN];
typedef unordered_map<LL, LL> mp;
mp cnt[];

int calc(int x) {
    ;
    ) {
        res++;
        x /= ;
    }
    return res;
}

int main() {
#ifndef ONLINE_JUDGE
    freopen("d.in", "r", stdin);
#endif // ONLINE_JUDGE
    ;
    p10[] = ;
    ; i <= ; ++i)p10[i] = p10[i - ] * 10LL;
    read(n), read(k);
    ; i < n; ++i) {
        read(a[i]);
        len[i] = calc(a[i]);
        ml = max(ml, (int)len[i]);
        cnt[len[i]][a[i] % k]++;
    }
    LL ans = ;
    ; j <= ml; ++j) {
        for(auto e : cnt[j]) {
            LL A = e.fi;
            ; i <= ml; ++i) {
                LL B = ((k - A * (p10[i] % k)) % k + k) % k;
                B = (B + k) % k;
                mp::iterator it2 = cnt[i].find(B);
                if(it2 != cnt[i].end())ans += e.se * it2->se;
            }
        }
    }
    ; i < n; ++i) {
        LL t2 = (((a[i] % k) * (p10[len[i]] % k) % k) + a[i]) % k;
        )ans--;
    }
    cout << ans << endl;
    ;
}
/*
2 27961
1000000000 1000000000
*/