codeforces 620D Professor GukiZ and Two Arrays
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1 #include <bits/stdc++.h> 2 3 using namespace std; 4 5 const int maxn = 2000 + 50; 6 7 const long long inf = 1e18; 8 9 int n, m; 10 11 long long suma, sumb; 12 13 int a[maxn], b[maxn]; 14 15 long long dbl_a[maxn], dbl_b[maxn]; 16 17 long long sum_dbl_a[maxn * maxn + 10], sum_dbl_b[maxn * maxn + 10]; 18 19 int main() 20 { 21 map<int, int> pos_a; 22 scanf("%d", &n); 23 for(int i = 1; i <= n; ++i){ 24 scanf("%d", &a[i]); 25 suma += a[i]; 26 dbl_a[i] = a[i] * 2; 27 pos_a[dbl_a[i]] = i; 28 } 29 30 map<long long, pair<int, int>> m_a; 31 map<int, long long> two_pos_a; 32 int cnt_a = 0, cnt_b = 0; 33 for(int i = 1; i <= n; ++i){ 34 for(int j = i + 1; j <= n; ++j){ 35 m_a[dbl_a[i] + dbl_a[j]] = make_pair(i, j); 36 sum_dbl_a[++cnt_a] = dbl_a[i] + dbl_a[j]; 37 } 38 } 39 scanf("%d", &m); 40 for(int i = 1; i <= m; ++i){ 41 scanf("%d", &b[i]); 42 sumb += b[i]; 43 dbl_b[i] = b[i] * 2; 44 } 45 46 long long diff = (suma - sumb); 47 48 if(diff == 0) { 49 printf("0\n0\n"); 50 return 0; 51 } 52 53 sort(dbl_a+1, dbl_a+1+n); 54 55 long long one_swap_ans = abs(suma - sumb); 56 long long two_swap_ans = abs(suma - sumb); 57 int one_swap_a = 0, one_swap_b = 0; 58 int two_swap_a_1 = 0, two_swap_a_2 = 0, two_swap_b_1 = 0, two_swap_b_2 = 0; 59 60 for(int i = 1; i <= m; ++i){ 61 long long tmp = diff + dbl_b[i]; 62 int pos = lower_bound(dbl_a+1, dbl_a+n+1, tmp) - dbl_a; 63 //printf("%d\n", pos); 64 long long big_res = inf; 65 if(pos <= n) big_res = dbl_a[pos-1]; 66 67 long long small_res = inf; 68 if(pos - 1 >= 1) small_res = dbl_a[pos - 1]; 69 70 if(abs(big_res - tmp) > abs(small_res - tmp)){ 71 //printf("%d %d %d\n",pos_a[10], dbl_a[pos-1], pos); 72 long long temp = abs(small_res - tmp); 73 if(one_swap_ans > temp){ 74 one_swap_ans = temp; 75 one_swap_a = pos_a[dbl_a[pos-1]], one_swap_b = i; 76 } 77 } else { 78 long long temp = abs(big_res - tmp); 79 if(one_swap_ans > temp){ 80 one_swap_ans = temp; 81 one_swap_a = pos_a[dbl_a[pos]], one_swap_b = i; 82 } 83 } 84 85 } 86 87 88 89 sort(sum_dbl_a + 1, sum_dbl_a + cnt_a + 1); 90 91 92 for(int i = 1; i <= m; ++i){ 93 for(int j = i + 1; j <= m; ++j){ 94 long long tmp = diff + dbl_b[i] + dbl_b[j]; 95 int pos = lower_bound(sum_dbl_a+1, sum_dbl_a+cnt_a+1, tmp) - sum_dbl_a; 96 long long big_res = inf; 97 if(pos <= cnt_a) big_res = sum_dbl_a[pos]; 98 99 long long small_res = inf; 100 if(pos - 1 >= 1) small_res = sum_dbl_a[pos - 1]; 101 102 if(abs(big_res - tmp) > abs(small_res - tmp)){ 103 long long temp = abs(small_res - tmp); 104 105 if(two_swap_ans > temp){ 106 two_swap_ans = temp; 107 two_swap_a_1 = m_a[sum_dbl_a[pos-1]].first; 108 two_swap_a_2 = m_a[sum_dbl_a[pos-1]].second; 109 two_swap_b_1 = i; 110 two_swap_b_2 = j; 111 } 112 } else { 113 long long temp = abs(big_res - tmp); 114 115 if(two_swap_ans > temp){ 116 two_swap_ans = temp; 117 two_swap_a_1 = m_a[sum_dbl_a[pos]].first; 118 two_swap_a_2 = m_a[sum_dbl_a[pos]].second; 119 120 two_swap_b_1 = i; 121 two_swap_b_2 = j; 122 } 123 } 124 125 } 126 } 127 128 129 long long ans = min(abs(diff), min(abs(one_swap_ans), abs(two_swap_ans))); 130 printf("%I64d\n", ans); 131 if(ans == diff){ 132 printf("0\n"); 133 } else if(one_swap_ans == ans){ 134 printf("1\n"); 135 printf("%d %d\n", one_swap_a, one_swap_b); 136 } else { 137 printf("2\n"); 138 printf("%d %d\n", two_swap_a_1, two_swap_b_1); 139 printf("%d %d\n", two_swap_a_2, two_swap_b_2); 140 } 141 142 return 0; 143 }
只交换一次的情况
原来 |Suma - Sumb|
交换后 Suma - xi + yj - (sumb - yj + xi) = suma - sumb + 2(yj - xi) -> 0
则在确定xi的情况下 需要在 b序列里找到 最接近 xi 的 (suma - sumb) /2 + yj的 yj
交换两次的情况
原来 |suma - sumb|
交换后 Suma - xi - xj + yk + yl - (sumb + xi + xj - yk - yl) = suma - sumb + 2((yk + yl) - (xi + xj)) -> 0
则在确定 xi xj的情况下 需要在b序列里找到 最接近 xi + xj的 (suma - sumb)/2 + yk + yl的 yk 和 yl
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