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  • Solution -「CF 1380F」Strange Addition

    (mathcal{Description})

      Link.

      定义两个数在进行加法时,进位单独作为一位。例如:

    .

      给定一个 (n) 为数和 (m) 次修改操作,每次修改会修改 (n) 位数的某一位数字。在每次修改后求出有多少对数以上述规则相加后的得数为这个 (n) 为数。

      (n,mle5 imes 10^5)

    (mathcal{Solution})

      显然的 DDP。

      令 (f_i) 表示加和为目标数后 (i) 为数字的数对数量。那么:

    [ w_{i+1} egin{pmatrix} f_i\ f_{i-1} end{pmatrix} = egin{pmatrix} f_{i+1}\ f_i end{pmatrix} ]

      其中 (w_i~(i=1,2,dots,18)) 分别表示每种“一位数”的转移矩阵,只有 (i=1) 需要特殊处理。

      具体看代码吧(摊手。

    (mathcal{Code})

    /* Clearink */
    
    #include <cstdio>
    
    inline int rint () {
    	int x = 0, f = 1; char s = getchar ();
    	for ( ; s < '0' || '9' < s; s = getchar () ) f = s == '-' ? -f : f;
    	for ( ; '0' <= s && s <= '9'; s = getchar () ) x = x * 10 + ( s ^ '0' );
    	return x * f;
    }
    
    template<typename Tp>
    inline void wint ( Tp x ) {
    	if ( x < 0 ) putchar ( '-' ), x = ~ x + 1;
    	if ( 9 < x ) wint ( x / 10 );
    	putchar ( x % 10 ^ '0' );
    }
    
    const int MAXN = 5e5, MOD = 998244353;
    const int w[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1 };
    int n, m;
    char num[MAXN + 5];
    
    inline int add ( int a, const int b ) { return ( a += b ) < MOD ? a : a - MOD; }
    inline int mul ( long long a, const int b ) { return ( a *= b ) < MOD ? a : a % MOD; }
    
    struct Matrix {
    	int mat[2][2];
    	Matrix (): mat {} {}
    	Matrix ( const int a, const int b, const int c, const int d ):
    		mat { a, b, c, d } {}
    
    	inline int* operator [] ( const int key ) { return mat[key]; }
    
    	inline Matrix operator * ( Matrix& t ) const {
    		Matrix ret;
    		for ( int i = 0; i < 2; ++ i ) {
    			for ( int k = 0; k < 2; ++ k ) {
    				for ( int j = 0; j < 2; ++ j ) {
    					ret[i][j] = add ( ret[i][j], mul ( mat[i][k], t[k][j] ) );
    				}
    			}
    		}
    		return ret;
    	}
    
    	inline void _show () const {
    #ifdef RYBY
    		for ( int i = 0; i < 2; ++ i ) {
    			for ( int j = 0; j < 2; ++ j ) {
    				printf ( "%d ", mat[i][j] );
    			}
    			putchar ( '
    ' );
    		}
    #endif
    	}
    };
    
    inline Matrix digit ( const int d ) {
    	return Matrix ( w[num[d] - '0'], num[d] ^ '1' || d == n ? 0 : w[10 + num[d + 1] - '0'], 1, 0 );
    }
    
    struct SegmentTree {
    	Matrix mat[MAXN * 2 + 5];
    
    	inline int id ( const int l, const int r ) { return ( l + r ) | ( l != r ); }
    
    	inline int build ( const int l, const int r ) {
    		int rt = id ( l, r ), mid = l + r >> 1;
    		if ( l == r ) return ( mat[rt] = digit ( l ) )._show (), rt;
    		int lc = build ( l, mid ), rc = build ( mid + 1, r );
    		return mat[rt] = mat[lc] * mat[rc], rt;
    	}
    
    	inline void update ( const int l, const int r, const int x, const char d ) {
    		int rt = id ( l, r ), mid = l + r >> 1;
    		if ( l == r ) return num[l] = d, mat[rt] = digit ( l ), void ();
    		if ( x <= mid ) update ( l, mid, x, d );
    		else update ( mid + 1, r, x, d );
    		mat[rt] = mat[id ( l, mid )] * mat[id ( mid + 1, r )];
    	}
    } segt;
    
    int main () {
    	n = rint (), m = rint ();
    	scanf ( "%s", num + 1 );
    	int rt = segt.build ( 1, n );
    	for ( int x, d; m --; ) {
    		x = rint (), d = rint ();
    		segt.update ( 1, n, x, d ^ '0' );
    		if ( x > 1 ) segt.update ( 1, n, x - 1, num[x - 1] );
    		segt.mat[rt]._show ();
    		wint ( segt.mat[rt][0][0] ), putchar ( '
    ' );
    	}
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/rainybunny/p/13740512.html
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