题目链接 http://acm.hdu.edu.cn/showproblem.php?pid=1104 在做这道题目一定要对同余定理有足够的了解，所以对这道...
下面直接上图吧，稍后解释关于mod和remainder以及负数求余求模的区别。 mod是模运算，remainder是求余运算，如果被除数是正整数，mod和remainder的结果没区别。... -14 ÷ 3 = -4 ··· -2（余数符号和被除数符号必须相同） 余数就是-2，这里如果是remainder取余运算，-14 remainder 3 = -2，结果是-2 如果是mod模运算...，这里如果是remainder取余运算，-10 remainder 3 = -1，结果是-1 如果是mod模运算，在remainder基础上判断正负，这里-1是负数，加上模数(可认为是除法的除数)，这里加上...3，mod模运算结果就是2，-10 mod 3 = 2 7 ÷ -3 = -2 ··· 1 余数是1，除数是负数，这里只能取模运算remiander 7 remainder -3 = 1，结果是1。...0%-5=0 0mod3=0 0remainder-3=0 ......
] == 1'b0) begin judge = remainder_r - divisor_move; end else begin judge = remainder_r...[WIDTH * 2 - 1:0] <= dividend; remainder_r[2 * WIDTH] <= 'b0; divisor_move[3...}) && (dout == 'b0)) begin //开始运算条件 remainder_r <= remainder_r; dout...[2 * WIDTH - 1] == 1'b1) begin //调整余数 remainder_r <= remainder_r + divisor_lock;...dout <= dout; end end end assign remainder = remainder_r[WIDTH - 1:0]; endmodule
to_string(quotient); remainder = remainder % ldenominator * 10;//得到余数并乘以10 }...remainder = remainder % ldenominator * 10; } return result.toString(); } } Python...map = {} while remainder !...= 0: if map.has_key(remainder): position = map[remainder]...) remainder = remainder % pdenominator * 10 return result
); 定义端口，其中remainder前WIDTH位均为0，可以不连接 reg [3 * WIDTH - 1:0]divisor_lock; reg [WIDTH - 1:0]divisor_ref;...= 'b0; end else if((dout == 'b0) && (remainder < divisor_lock)) begin remainder <= remainder...divisor_lock) begin remainder <= remainder - divisor_lock; dout <= {dout[2 *...WIDTH - 2:0],1'b1}; end else begin remainder <= remainder; dout <= {dout...[2 * WIDTH - 2:0],1'b0}; end end else begin {remainder,dout} <= {remainder,dout};
= 0: remainder = num % n if 20 > remainder > 9: remainder_string = baseStr...[remainder] elif remainder >=20: remainder_string = "("+str(remainder)+")" ...else: remainder_string = str(remainder) new_num_str = remainder_string+new_num_str
, output reg [WIDTH - 1:0]this_dout, output reg [2 * WIDTH - 1:0]remainder_dout ); wire [2...* WIDTH - 1:0]target_data = {remainder_din[2 * WIDTH - 3:0],radicand[2 * STEP +:2]}; wire [2 * WIDTH...remainder_din = square[i + 1].remainder_dout; assign last_dout = square[i + 1].this_dout...(remainder_din), .this_dout(this_dout), .remainder_dout(remainder_dout)...); end endgenerate assign dout = square[0].this_dout; assign remainder = square[0].remainder_dout
); reg [2 * WIDTH:0]remainder_r; reg [3 * WIDTH - 1:0]divisor_move; reg [WIDTH - 1:0]divisor_lock;...always @ (posedge clk or negedge rst_n) begin if(~rst_n) begin //初始化 {remainder_r...}) && (dout == 'b0)) begin //开始条件 remainder_r <= remainder_r; dout...[2 * WIDTH] == 1'b0) begin //执行减法 remainder_r <= remainder_r - divisor_move;...dout <= dout; end end end assign remainder = remainder_r[WIDTH - 1:0]; endmodule
It is used to calculate the remainder and quotient, this function is the same as the remainder() function...It accepts three parameters (numerator, denominator, and quotient) and returns the remainder, assigns...); cout << "numerator : " << x << endl; cout << "denominator: " << y << endl; cout << "remainder... : 0 quotient : 5 numerator : 10.34 denominator: 2.5 remainder : 0.34 quotient : 4 numerator ...: -10.02 denominator: 2.3 remainder : -0.82 quotient : -4 numerator : 10.23 denominator: -2 remainder
<< STEP); always @ (posedge clk or negedge rst_n) begin if(~rst_n) begin {remainder_dout...= sub; end else begin remainder_dout = remainder_din; end quotient...if(remainder_din[3 * WIDTH] == 'b0) begin remainder_dout = sub; quotient = ~(...; always @ (*) begin if(remainder_final[3 * WIDTH] == 1'b0) begin remainder = remainder_final...[WIDTH - 1:0]; end else begin remainder = remainder_final[WIDTH - 1:0] + divisor; end
, output reg [WIDTH - 1:0]this_dout, output reg [2 * WIDTH:0]remainder_dout ); wire [2 * WIDTH...:0]target_data = {remainder_din[2 * WIDTH],remainder_din[2 * WIDTH - 3:0],radicand[2 * STEP +:2]}; wire...remainder_din = square[i + 1].remainder_dout; assign last_dout = square[i + 1].this_dout...(remainder_din), .this_dout(this_dout), .remainder_dout(remainder_dout)...); end endgenerate assign dout = square[0].this_dout; // assign remainder = square[0].remainder_dout
否则，将遍历数组，求前缀和，前缀和与p的余数设为remainder，题目需要移除的部分是两个前缀和相减后的部分，其与p的余数应为k，即(remainder - temp) % p == k，temp是需要寻找的余数...，由于remainder - temp为负数时，公式变为p + remainder - temp = k，通过公式变换，可以得到temp = p + remainder - k，由于remainder...- temp大于0时不需要加上p，因此公式变为temp = (remainder + p - k) % p。...= total % p temp = (remainder + p - k) % p if temp in stat:...length = min(length, i - stat[temp]) stat[remainder] = i if length == len(nums):
= k % n; vector v(n); for(int i = 0; i < remainder; i++) { v[i] = nums...[i + start]; } for(int i = 0; i < start; i++) { v[i + remainder] = nums[i...= k % n; vector v(remainder); for(int i = 0; i < remainder; i++) {...v[i] = nums[i + start]; } for(int i = n - 1; i >= remainder; i--) { nums[...i] = nums[i - remainder]; } for(int i = 0; i < remainder; i++) { nums[i]
int reverse(int x) { if(x == 0 || x == INT_MIN) { return 0; } int remainder...= 0; int y = abs(x); queue number; while(y) { remainder = y...% 10; number.push(remainder); y /= 10; } int result = 0;...= 0; while(y) { remainder = y % 10; if((INT_MAX - remainder) / 10 <...result) { return 0; } result = result * 10 + remainder;
digits { mod := digits[i] % 3 remainder[mod] = append(remainder[mod], digits[i]) }...// 在可拿的情况下，保证拿的都是最大的 sort.Ints(remainder[1]) sort.Ints(remainder[2]) resDigits := remainder...[0] // 以余 0 为基础 len1, len2 := len(remainder[1]), len(remainder[2]) // 尽可能的多拿，[1,1,1,2]...remainder[2][0] = -1 remainder[2][1] = -1 } } else { if len2 > 0 { remainder...[2][0] = -1 } else if len1 > 1 { remainder[1][0] = -1 remainder[1][1] = -1 }
integers N equal parts sum # 拆分整数 def split_integer(m, n): assert n > 0 quotient = int(m / n) remainder...= m % n if remainder > 0: return [quotient] * (n - remainder) + [quotient + 1] * remainder...if remainder < 0: return [quotient - 1] * -remainder + [quotient] * (n + remainder)
---- it_works stdout ---- 1-remainder = "a b c d e f" 1-remainder = "b c d e f" 1-remainder = "c d e...f" 1-remainder = "d e f" 1-remainder = "e f" 3-remainder = "f" 2-remainder = "" ---- tail_test stdout...---- 1-remainder = "a b c d " 1-remainder = "b c d " 1-remainder = "c d " 1-remainder = "d " 2-remainder...= self.remainder.as_mut()?...&remainder[..next_delim]; *remainder = &remainder[(next_delim + self.delimiter.len())..];
= divmod(number, 2) bits.insert(0, remainder) return bits else: return...while number: 15:29:11.327032 line 8 number, remainder = divmod(number, 2) Modified...(0, remainder) Modified var:.. bits = [1, 0] 15:29:11.327032 line 7 while number: 15:...29:11.327032 line 8 number, remainder = divmod(number, 2) Modified var:.. number...= 0 15:29:11.327032 line 9 bits.insert(0, remainder) Modified var:.. bits = [1, 1
/ divisor; remainder = dividend % divisor; } void divideWithPointers(int dividend, int divisor, int...* quotient, int* remainder) { if (quotient) *quotient = dividend / divisor; if (remainder) *remainder...) = divide(14, 3); std::cout << quotient << ", " << remainder << std::endl; 3....auto divide(int dividend, int divisor) { struct result { int quotient; int remainder; };...) { quotient_out.emplace(dividend / divisor); remainder_out.emplace(dividend % divisor); } 6.
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