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| 1 | +package weekly; |
| 2 | + |
| 3 | +import java.text.ParseException; |
| 4 | +import java.text.SimpleDateFormat; |
| 5 | +import java.time.LocalTime; |
| 6 | +import java.util.ArrayList; |
| 7 | +import java.util.Arrays; |
| 8 | +import java.util.Date; |
| 9 | +import java.util.HashMap; |
| 10 | +import java.util.List; |
| 11 | +import java.util.Map; |
| 12 | +import java.util.PriorityQueue; |
| 13 | + |
| 14 | +public class wk316 { |
| 15 | + |
| 16 | + //ranking: 942 / 6387 |
| 17 | + |
| 18 | + //又是这种时间题 |
| 19 | + public boolean haveConflict(String[] event1, String[] event2) { |
| 20 | + |
| 21 | + return event1[0].compareTo(event2[1]) <= 0 && event1[1].compareTo(event2[0]) >= 0; |
| 22 | + |
| 23 | + } |
| 24 | + |
| 25 | +// boolean check(LocalTime a, LocalTime b, LocalTime c, LocalTime d) { |
| 26 | +// if (b.compareTo(c) >= 0 && b.compareTo(d) <= 0) return true; |
| 27 | +// if (d.compareTo(a) >= 0 && d.compareTo(b) <= 0) return true; |
| 28 | +// return false; |
| 29 | +// } |
| 30 | + |
| 31 | + |
| 32 | + //暴力写的 |
| 33 | + public int subarrayGCD(int[] nums, int k) { |
| 34 | + int ans = 0; |
| 35 | + for (int i = 0; i < nums.length; i++) { |
| 36 | + int g = 0; |
| 37 | + for (int j = i; j < nums.length; j++) { |
| 38 | + g = gcd(g, nums[j]); |
| 39 | + if (g == k) ans += 1; |
| 40 | + if (g % k > 0) break; |
| 41 | + } |
| 42 | + } |
| 43 | + return ans; |
| 44 | + } |
| 45 | + |
| 46 | + int gcd(long a, long b) { |
| 47 | + while (b != 0) { |
| 48 | + long temp = a % b; |
| 49 | + a = b; |
| 50 | + b = temp; |
| 51 | + } |
| 52 | + return (int) a; |
| 53 | + } |
| 54 | + |
| 55 | + |
| 56 | + //前缀和容易考虑一点, 但也可以直接考虑sum |
| 57 | + /* public long minCost(int[] nums, int[] cost) { |
| 58 | + long[][] pack = new long[nums.length][2]; |
| 59 | + for (int i = 0; i < nums.length; i++) { |
| 60 | + pack[i] = new long[]{nums[i], cost[i]}; |
| 61 | + } |
| 62 | + Arrays.sort(pack, (a, b) -> Long.compare(a[0], b[0])); |
| 63 | +
|
| 64 | + long res = 0; |
| 65 | + for (int i = 1; i < pack.length; i++) { |
| 66 | + res += (pack[i][0] - pack[0][0]) * pack[i][1]; |
| 67 | + } |
| 68 | + for (int i = 1; i < pack.length; i++) { |
| 69 | + pack[i][1] += pack[i - 1][1]; |
| 70 | + } |
| 71 | + long ans = Long.MAX_VALUE; |
| 72 | + for (int i = 1; i < pack.length; i++) { |
| 73 | + res += (pack[i][0] - pack[i - 1][0]) * (pack[i - 1][1] - (pack[nums.length - 1][1] - pack[i - 1][1])); |
| 74 | + ans = Math.min(ans, res); |
| 75 | + } |
| 76 | + return ans; |
| 77 | + }*/ |
| 78 | + //不用前缀和 |
| 79 | + public long minCost(int[] nums, int[] cost) { |
| 80 | + long[][] pack = new long[nums.length][2]; |
| 81 | + for (int i = 0; i < nums.length; i++) { |
| 82 | + pack[i] = new long[]{nums[i], cost[i]}; |
| 83 | + } |
| 84 | + Arrays.sort(pack, (a, b) -> Long.compare(a[0], b[0])); |
| 85 | + |
| 86 | + long res = 0; |
| 87 | + long sumCost=pack[0][1]; |
| 88 | + for (int i = 1; i < pack.length; i++) { |
| 89 | + res += (pack[i][0] - pack[0][0]) * pack[i][1]; |
| 90 | + sumCost+=pack[i][1]; |
| 91 | + } |
| 92 | + long ans = res; |
| 93 | + for (int i = 1; i < pack.length; i++) { |
| 94 | + sumCost-=2*pack[i-1][1]; |
| 95 | + res -= (pack[i][0] - pack[i - 1][0]) * sumCost; |
| 96 | + ans = Math.min(ans, res); |
| 97 | + } |
| 98 | + return ans; |
| 99 | + } |
| 100 | + |
| 101 | + |
| 102 | + //分奇数偶数考虑,要分很多情况 |
| 103 | + /* public long makeSimilar(int[] nums, int[] target) { |
| 104 | + List<Integer> nums1 = new ArrayList<>(); |
| 105 | + List<Integer> nums2 = new ArrayList<>(); |
| 106 | + List<Integer> target1 = new ArrayList<>(); |
| 107 | + List<Integer> target2 = new ArrayList<>(); |
| 108 | + Arrays.sort(nums); |
| 109 | + Arrays.sort(target); |
| 110 | + for (int i = 0; i < nums.length; i++) { |
| 111 | + if (nums[i] % 2 == 1) { |
| 112 | + nums1.add(nums[i]); |
| 113 | + } else { |
| 114 | + nums2.add(nums[i]); |
| 115 | + } |
| 116 | + if (target[i] % 2 == 1) { |
| 117 | + target1.add(target[i]); |
| 118 | + } else { |
| 119 | + target2.add(target[i]); |
| 120 | + } |
| 121 | + } |
| 122 | + long[] a = help(nums1, target1, 0, 0); |
| 123 | + long[] b = help(nums2, target2, a[1], a[2]); |
| 124 | + System.out.println(a + " " + b); |
| 125 | + return a[0] + b[0]; |
| 126 | + } |
| 127 | +
|
| 128 | + long[] help(List<Integer> nums, List<Integer> target, long up, long down) { |
| 129 | + long count = 0; |
| 130 | + for (int i = 0; i < nums.size(); i++) { |
| 131 | + int a = nums.get(i), b = target.get(i); |
| 132 | + if (a > b) { |
| 133 | + int need = (a - b) / 2; |
| 134 | + if (down >= need) { |
| 135 | + down -= need; |
| 136 | + } else { |
| 137 | + count += (need - down); |
| 138 | + up += (need - down); |
| 139 | + down = 0; |
| 140 | + } |
| 141 | +
|
| 142 | + } else if (a < b) { |
| 143 | + int need = (b - a) / 2; |
| 144 | + if (up >= need) { |
| 145 | + up -= need; |
| 146 | + } else { |
| 147 | + count += (need - up); |
| 148 | + down += (need - up); |
| 149 | + up = 0; |
| 150 | + } |
| 151 | + } |
| 152 | + } |
| 153 | + return new long[]{count, up, down}; |
| 154 | + }*/ |
| 155 | + |
| 156 | + |
| 157 | + //大佬的做法,把奇数变成负数,区分奇偶,然后计算绝对值差,最后/4即可,因为题目说一定有答案 |
| 158 | + public long makeSimilar(int[] nums, int[] target) { |
| 159 | + f(nums); |
| 160 | + f(target); |
| 161 | + long ans = 0L; |
| 162 | + for (int i = 0; i < nums.length; ++i) |
| 163 | + ans += Math.abs(nums[i] - target[i]); |
| 164 | + return ans / 4; |
| 165 | + } |
| 166 | + |
| 167 | + private void f(int[] a) { |
| 168 | + // 由于元素都是正数,把奇数变成相反数,这样排序后奇偶就自动分开了 |
| 169 | + for (int i = 0; i < a.length; ++i) |
| 170 | + if (a[i] % 2 != 0) a[i] = -a[i]; |
| 171 | + Arrays.sort(a); |
| 172 | + } |
| 173 | + |
| 174 | +} |
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