-
前缀和+差分+ 树状数组
前缀和相关题：
「USACO16JAN」子共七 Subsequences Summing to Sevens
定理：若两个数相减 mod k == 0,那么这两个数mod k 的余数相同。（看别人的题解得到的）
#includeusing namespace std; const int N = 5e4; int n; int a[N + 1];//前n项和mod 7 int bgin[7],tail[7]; //bgin存第1个mod为i的数据编号，tail存最后一个mod为i的数据编号 int main() { // 若两个数相减mod 7==0，那么这俩个数mod 7的余数一定相同 ios::sync_with_stdio(false); int res = 0;//存结果 cin >> n ; for(int i = 1; i <= n; i ++) { cin >> a[i]; a[i] += a[i-1]; a[i] %= 7; } for(int i = 0; i <= n; i ++) //从前往后搜，且这里必须是i=0开始，不能从i=1开始，考虑数据1 3 4即明白 tail[a[i]] = i; for(int i = n; i >= 0; i --) //从后往前搜， 并且这里必须要包含0也即i>=0 bgin[a[i]] = i; for(int i = 0; i < 7; i ++) res = max(res, tail[i] - bgin[i]); cout << res << endl; return 0; } - 1
- 2
- 3
- 4
- 5
- 6
- 7
- 8
- 9
- 10
- 11
- 12
- 13
- 14
- 15
- 16
- 17
- 18
- 19
- 20
- 21
- 22
- 23
- 24
- 25
- 26
- 27
- 28
- 29
- 30
- 31
- 32
- 33
- 34
- 35
公式推到：（假设数据有序）
第i个位置的牛与前i-1个牛聊天音量的总和：
x i − x i − 1 + x i − x i − 2 + ⋯ + x i − x 2 + x i − x 1 = ( i − 1 ) ⋅ ∑ k = 1 i − 1 x k x_i - x_{i-1} + x_i - x_{i-2} +\cdots+ x_i -x_2+x_i - x_1 = (i-1)\cdot\sum\limits_{k=1}^{i-1}x_k xi−xi−1+xi−xi−2+⋯+xi−x2+xi−x1=(i−1)⋅k=1∑i−1xk
下式res结果要乘2 是因为只计算了 ( x i > x j ) 时 x i − x j (x_i>x_j)时x_i-x_j (xi>xj)时xi−xj并没有计算 x j − x i x_j-x_i xj−xi，且 ∣ x i − x j ∣ = ∣ x j − x i ∣ \lvert x_i-x_j\rvert=\lvert x_j-x_i\rvert ∣xi−xj∣=∣xj−xi∣。r e s = 2 ⋅ ∑ i = 1 n ( i − 1 ) ⋅ ∑ k = 1 i − 1 x k res =2\cdot \sum\limits_{i=1}^{n}(i-1)\cdot\sum\limits_{k=1}^{i-1}x_k res=2⋅i=1∑n(i−1)⋅k=1∑i−1xk
#includeusing namespace std; const int N = 1e5 + 1; typedef long long ll; ll n; ll a[N]; //存放数字 ll sum[N]; //前n项和 int main() { ios::sync_with_stdio(false); cin >> n; ll res = 0;//结果 for(int i = 1; i <= n; i ++) cin >> a[i]; //排序 sort(a+1,a+n+1); //求前n项和 for(int i = 1; i <= n; i ++) sum[i] = sum[i-1] + a[i]; // 求最终结果 for(int i = 1; i <= n; i ++) res += (i - 1)*a[i] - sum[i-1];//此公式需要推到 cout << res * 2 << endl; } - 1
- 2
- 3
- 4
- 5
- 6
- 7
- 8
- 9
- 10
- 11
- 12
- 13
- 14
- 15
- 16
- 17
- 18
- 19
- 20
- 21
- 22
- 23
- 24
- 25
- 26
- 27
二维前缀和：
#includeusing namespace std; const int N = 100 + 3; int n, m; int matrix[N][N]; int sum[N][N]; //前缀和 int main() { ios::sync_with_stdio(false); int res = 0; cin >> n >> m; for(int i = 1; i <= n; i ++) for(int j = 1; j <= m; j ++) { cin >> matrix[i][j]; sum[i][j] = sum[i - 1][j] + sum[i][j - 1] - sum[i - 1][j - 1] + matrix[i][j]; } for(int len = 1; len <= min(n, m); len ++)//边长 for(int i = len; i <= n; i ++) for(int j = len; j <= m; j ++) { if((sum[i][j] - sum[i - len][j] - sum[i][j - len] + sum[i - len][j - len]) == (len * len)) res = max(res, len); } cout << res << endl; return 0; } - 1
- 2
- 3
- 4
- 5
- 6
- 7
- 8
- 9
- 10
- 11
- 12
- 13
- 14
- 15
- 16
- 17
- 18
- 19
- 20
- 21
- 22
- 23
- 24
- 25
- 26
- 27
- 28
- 29
- 30
- 31
- 32
以下都是AC代码，为了防止MLE进行了空间优化，也即sum即存储原始数据，也存储前缀和。
#includeusing namespace std; const int N = 5e3 + 3; int n, m; int sum[N][N];//前缀和 int main() { ios::sync_with_stdio(false); int res = 0, mx = 0, my = 0; cin >> n >> m; for(int i = 0; i < n; i ++) { int x, y, v; cin >> x >> y >> v; sum[x + 1][y + 1] = v; mx = max({mx, x + 1, m}); my = max({my, y + 1, m});//必须把m算进去，考虑此类情况：n=1 m= 2 ,第二行输入 0,0 ,1 } for(int i = 1; i <= mx; i ++) for(int j = 1; j <= my; j ++) sum[i][j] = sum[i - 1][j] + sum[i][j - 1] - sum[i - 1][j - 1] + sum[i][j]; for (int i = m; i <= mx; i ++) for(int j = m; j <= my; j ++) res = max(res, sum[i][j] - sum[i - m][j] - sum[i][j - m] + sum[i - m][j - m]); cout << res << endl; return 0; } - 1
- 2
- 3
- 4
- 5
- 6
- 7
- 8
- 9
- 10
- 11
- 12
- 13
- 14
- 15
- 16
- 17
- 18
- 19
- 20
- 21
- 22
- 23
- 24
- 25
- 26
- 27
- 28
- 29
- 30
- 31
- 32
#includeusing namespace std; const int N = 5000 + 11; int n, m; int sum[N][N];//前缀和 int main() { ios::sync_with_stdio(false); int res = 0, mx = 0, my = 0; cin >> n >> m; for(int i = 0; i < n; i ++) { int x, y, v; cin >> x >> y >> v; sum[x + 1][y + 1] = v;//这个是让坐标从1开始，而不是从0开始 mx = max(mx, x + 1); my = max(my, y + 1); } for(int i = 1; i <= min(mx + m, 5001); i ++) //这里i不能小于等于N，因为你数组就开N个， 当然如果这里为了省事，可直接i<=5001 下面一样 for(int j = 1; j <= min(my + m, 5001); j ++) sum[i][j] = sum[i - 1][j] + sum[i][j - 1] - sum[i - 1][j - 1] + sum[i][j];//这里其实进行了空间的优化 for (int i = m; i <= min(mx + m, 5001); i ++) for(int j = m; j <= min(my + m, 5001); j ++) res = max(res, sum[i][j] - sum[i - m][j] - sum[i][j - m] + sum[i - m][j - m]); cout << res << endl; return 0; } - 1
- 2
- 3
- 4
- 5
- 6
- 7
- 8
- 9
- 10
- 11
- 12
- 13
- 14
- 15
- 16
- 17
- 18
- 19
- 20
- 21
- 22
- 23
- 24
- 25
- 26
- 27
- 28
- 29
- 30
- 31
- 32
- 33
#includeusing namespace std; const int N = 600 + 3; int n, l, r, t; int sum[N][N]; int main() { ios::sync_with_stdio(false); int res = 0; cin >> n >> l >> r >> t; for(int i = 1; i <= n; i ++) for(int j = 1; j <= n; j ++) { cin >> sum[i][j]; sum[i][j] = sum[i-1][j] + sum[i][j - 1] - sum[i - 1][j - 1] + sum[i][j]; } // 中心点为参考 for(int i = 1; i <= n; i ++) for(int j = 1; j <= n; j ++) { int loc_l = max(1, j - r); int loc_r = min(n, j + r);//不能直接把变量设为r,t,会与前面的变量名重复。 int loc_t = max(1, i - r); int loc_b = min(n, i + r); double counts = (loc_r - loc_l + 1) * (loc_b - loc_t + 1); //个数 if( ((sum[loc_b][loc_r] - sum[loc_b][loc_l - 1] - sum[loc_t - 1][loc_r] + sum[loc_t - 1][loc_l - 1]) / counts) <= t) res += 1; } cout << res << endl; return 0; } - 1
- 2
- 3
- 4
- 5
- 6
- 7
- 8
- 9
- 10
- 11
- 12
- 13
- 14
- 15
- 16
- 17
- 18
- 19
- 20
- 21
- 22
- 23
- 24
- 25
- 26
- 27
- 28
- 29
- 30
- 31
- 32
- 33
- 34
- 35
#includeusing namespace std; const int M = 1e5 + 1; int m; struct Node { int y,result; bool operator < (const struct Node n)//根据y排序 { if (y < n.y) return true; return false; } }node[M]; int sum[M]; int main() { ios::sync_with_stdio(false); int res = 0, maxTotal = 0;//res：最大阈值，maxTotal: 预测正确次数最大 cin >> m; for(int i = 1; i <= m; i ++) cin >> node[i].y >> node[i].result; sort(node + 1, node + m + 1); for(int i = 1; i <= m; i ++) sum[i] = sum[i-1] + node[i].result; for(int i = 1; i <= m; i ++) { if(node[i].y == node[i-1].y) //这一步是必须的，如果y连续出现相同的数，只需计算第一个，后面的不要再计算 ，因为下面的sum[i-1]会出错 continue; int total = (i - 1 - sum[i-1]) + (sum[m] - sum[i - 1]);//计算预测正确次数， +前部分是 小于y的预测正确的次数，+ 后半部分是 大于y预测正确的次数 //找最大的maxtotal和阈值 if(maxTotal <= total) { maxTotal = total; res = node[i].y; } } cout << res << endl; return 0; } - 1
- 2
- 3
- 4
- 5
- 6
- 7
- 8
- 9
- 10
- 11
- 12
- 13
- 14
- 15
- 16
- 17
- 18
- 19
- 20
- 21
- 22
- 23
- 24
- 25
- 26
- 27
- 28
- 29
- 30
- 31
- 32
- 33
- 34
- 35
- 36
- 37
- 38
- 39
- 40
- 41
- 42
- 43
- 44
- 45
- 46
- 47
- 48
- 49
- 50
- 51
- 52
- 53
- 54
差分相关题：
a i + 1 = b i + 1 − b i a_{i+1} = b_{i+1} - b_i ai+1=bi+1−bi其中 a a a是差分矩阵， b b b是原矩阵，对差分矩阵求前缀和即是原矩阵。如果对原矩阵 [ l , r ] [l,r] [l,r]增加k，那么可以对差分矩阵 a [ l ] + = k , a [ r + 1 ] − = k a[l] += k, a[r + 1] -= k a[l]+=k,a[r+1]−=k。可以减少时间复杂度。
∑ i = 1 n a i = ∑ i = 1 n ∑ j = 1 i b j = ∑ i = 1 n ( n − i + 1 ) b i = ( n + 1 ) ∑ i = 1 n b i − ∑ i = 1 n i ⋅ b i \sum\limits_{i=1}^n a_i = \sum\limits_{i=1}^n \sum\limits_{j=1}^i b_j = \sum\limits_{i=1}^n (n - i + 1) b_i = (n + 1)\sum\limits_{i =1}^n b_i-\sum\limits_{i = 1}^n i\cdot b_i i=1∑nai=i=1∑nj=1∑ibj=i=1∑n(n−i+1)bi=(n+1)i=1∑nbi−i=1∑ni⋅bi
有关 ∑ i = 1 n ∑ j = 1 i b j = ∑ i = 1 n ( n − i + 1 ) b i \sum\limits_{i=1}^ n \sum\limits_{j=1}^ i b_j = \sum\limits_{i=1}^n (n - i + 1) b_i i=1∑nj=1∑ibj=i=1∑n(n−i+1)bi 这一步的产生，考虑有n个 b 1 b_1 b1,n-1个 b 2 b_2 b2 ,以此类推有 n − i + 1 n-i+1 n−i+1个 b i b_i bi故得出结果。q + k ≤ t i ≤ q + k + c i − 1 q + k \le t_i \le q +k+c_i -1 q+k≤ti≤q+k+ci−1 则 t i + 1 − c i − k ≤ q ≤ t i − k t_i + 1- c_i - k \leq q \leq t_i - k ti+1−ci−k≤q≤ti−k,也即做核酸的时间点要在 [ t i + 1 − c i − k , t i − k ] [t_i + 1- c_i - k , t_i - k] [ti+1−ci−k,ti−k]内，对每个出行计划其范围内的时间点+1，时间点为整数，最终如果求q时刻满足出行的计划，直接根据下标q即可得。
#includeusing namespace std; const int N = 2e5 + 3; int n, m, k; int diff[N]; // 差分矩阵 ，差分矩阵的原矩阵表示在每个时间点，如果这个时间做核酸可以出行，则这里的值+1 int q[N]; int main() { ios::sync_with_stdio(false); cin >> n >> m >> k; for(int i = 0; i < n; i ++) { int ti, ci; cin >> ti >> ci; diff[max(1,ti + 1 - ci - k)] ++;//这里要和 1进行比较，如果不加会出现错误。 diff[max(1,ti - k + 1)] --; } for(int i = 1; i < N; i ++) diff[i] = diff[i - 1] + diff[i]; for(int i = 0; i < m; i ++) cin >> q[i]; for(int i = 0; i < m; i ++) cout << diff[q[i]] << endl; return 0; } - 1
- 2
- 3
- 4
- 5
- 6
- 7
- 8
- 9
- 10
- 11
- 12
- 13
- 14
- 15
- 16
- 17
- 18
- 19
- 20
- 21
- 22
- 23
- 24
- 25
- 26
- 27
- 28
P3397 地毯此题是很经典的二维差分问题。有关二维差分参见此博客
#includeusing namespace std; const int N = 1000 + 3; int n, m; int diff[N][N]; int main() { ios::sync_with_stdio(false); cin >> n >> m; for(int i = 0; i < m; i ++) { int lx, ly, rx, ry; cin >> lx >> ly >> rx >> ry; diff[lx][ly] ++; diff[lx][ry + 1] --; diff[rx + 1][ly] --; diff[rx + 1][ry + 1] ++; } //求前缀和 for(int i = 1; i <= n; i ++) for(int j = 1; j <= n; j ++) { diff[i][j] = diff[i-1][j] + diff[i][j - 1] - diff[i - 1][j - 1] + diff[i][j];//这里是根据差分求原矩阵 } //输出 for(int i = 1; i <= n; i ++) { for(int j = 1; j <= n; j ++) { cout << diff[i][j] << " "; } cout << endl; } return 0; } - 1
- 2
- 3
- 4
- 5
- 6
- 7
- 8
- 9
- 10
- 11
- 12
- 13
- 14
- 15
- 16
- 17
- 18
- 19
- 20
- 21
- 22
- 23
- 24
- 25
- 26
- 27
- 28
- 29
- 30
- 31
- 32
- 33
- 34
- 35
- 36
- 37
- 38
- 39
- 40
- 41
P4552 [Poetize6] IncDec Sequence
题解实在是不知道该怎么说，洛谷有相关题解，第一个写的很好。#includeusing namespace std; typedef long long LL; const int N = 100000 + 3; LL n; LL diff[N];//a是原矩阵，diff是差分矩阵 ,必须是LL，int会报错，2^31，需要开LL int main() { ios::sync_with_stdio(false); LL count_operator = 0, count_result = 0; cin >> n; LL last = 0; for(int i = 1; i <= n; i ++) { LL t; cin >> t; diff[i] = t - last; last = t; } LL sum_p = 0,sum_n = 0; for(int i = 2; i <= n; i ++) { if(diff[i] < 0) sum_n += diff[i]; else if(diff[i] > 0) sum_p += diff[i]; } count_operator = abs(sum_p + sum_n) + min(sum_p, abs(sum_n)); count_result = abs(sum_p + sum_n) + 1; cout << count_operator << endl; cout << count_result << endl; return 0; } - 1
- 2
- 3
- 4
- 5
- 6
- 7
- 8
- 9
- 10
- 11
- 12
- 13
- 14
- 15
- 16
- 17
- 18
- 19
- 20
- 21
- 22
- 23
- 24
- 25
- 26
- 27
- 28
- 29
- 30
- 31
- 32
- 33
- 34
- 35
- 36
- 37
- 38
- 39
树状数组
/* 单点修改，区间查询 2022.9.9 计算机中，负数用补码表示 ，所以lowbit = x & -x */ #includeusing namespace std; typedef long long ll; const ll N = 1e6 + 3; ll t[N]; //存储的原数组的树状数组。 struct Node { ll id, l, r; }node[N]; ll n, q; //单点修改,对下标为i的原数组增加x void add(ll x, ll k) { for(;x <= n; x += x & -x) t[x] += k; } // 求下标x的原数组前缀和 ll prefix(ll x) { ll ans = 0; for(;x >= 1; x -= x & -x) ans += t[x]; return ans; } // 求区间原数组[l,r]的区间和 ll quary(ll l, ll r) { return prefix(r) - prefix(l - 1); } int main() { cin >> n >> q; for(ll i = 1; i <= n; i ++) { ll temp; cin >> temp; add(i, temp); } for(ll i = 0; i < q; i ++) cin >> node[i].id >> node[i].l >> node[i].r; for(ll i = 0; i < q; i ++) { if(node[i].id == 1) add(node[i].l, node[i].r); else if(node[i].id == 2) cout << quary(node[i].l, node[i].r) << endl; } return 0; } - 1
- 2
- 3
- 4
- 5
- 6
- 7
- 8
- 9
- 10
- 11
- 12
- 13
- 14
- 15
- 16
- 17
- 18
- 19
- 20
- 21
- 22
- 23
- 24
- 25
- 26
- 27
- 28
- 29
- 30
- 31
- 32
- 33
- 34
- 35
- 36
- 37
- 38
- 39
- 40
- 41
- 42
- 43
- 44
- 45
- 46
- 47
- 48
- 49
- 50
- 51
- 52
- 53
- 54
- 55
- 56
- 57
- 58
- 59
写成模板
/* 单点修改，区间查询 2022.9.9 计算机中，负数用补码表示 ，所以lowbit = x & -x */ #includeusing namespace std; typedef long long ll; //写成模板 const ll N = 1e6 + 3; struct BIT { ll t[N], n; //t[n]存储的原数组的树状数组,n是元素组的元素个数。 //单点修改,对下标为i的原数组增加x inline void add(ll x, ll k) { for(;x <= n; x += x & -x) t[x] += k; } // 求下标x的原数组前缀和 inline ll prefix(ll x) { ll ans = 0; for(;x >= 1; x -= x & -x) ans += t[x]; return ans; } // 求区间原数组[l,r]的区间和 ，区间查询 inline ll quary(ll l, ll r) { return prefix(r) - prefix(l - 1); } }; struct BIT bit; struct Node { ll id, l, r; }node[N]; ll q; int main() { ios::sync_with_stdio(false); cin >> bit.n >> q; for(ll i = 1; i <= bit.n; i ++) { ll temp; cin >> temp; bit.add(i, temp); } for(ll i = 0; i < q; i ++) cin >> node[i].id >> node[i].l >> node[i].r; for(ll i = 0; i < q; i ++) { if(node[i].id == 1) bit.add(node[i].l, node[i].r); else if(node[i].id == 2) cout << bit.quary(node[i].l, node[i].r) << endl; } return 0; } - 1
- 2
- 3
- 4
- 5
- 6
- 7
- 8
- 9
- 10
- 11
- 12
- 13
- 14
- 15
- 16
- 17
- 18
- 19
- 20
- 21
- 22
- 23
- 24
- 25
- 26
- 27
- 28
- 29
- 30
- 31
- 32
- 33
- 34
- 35
- 36
- 37
- 38
- 39
- 40
- 41
- 42
- 43
- 44
- 45
- 46
- 47
- 48
- 49
- 50
- 51
- 52
- 53
- 54
- 55
- 56
- 57
- 58
- 59
- 60
- 61
- 62
- 63
/* time: 2022.9.9 区间修改，单点查询 */ #includeusing namespace std; typedef long long ll; const ll N = 1e6 + 3; struct BIT { ll n, t[N]; //t[N]存储的是原数组的差分数组的树状数组 //单点修改 （这里的单点是指对差分数组中的一个元素修改，而非原数组） inline void add(ll x,ll k) { for(; x<= n; x += x & -x) t[x] += k; } //区间修改,对原数组的[l,k]+k（指的是对原数组进行区间修改） inline void inter_add(ll l, ll r, ll k) { add(l, k); add(r + 1, -k); } //单点查询,(这里的单点查询，是对原数组求其某下标的值),其实是求差分数组的前缀和 inline ll ask(ll x) { ll ans = 0; for(;x >= 1; x -= x & -x) ans += t[x]; return ans; } }; struct BIT bit; ll q; struct Qu { ll id,l,r,x; }Q[N]; int main() { cin >> bit.n >> q; ll last = 0; for(ll i = 1; i <= bit.n; i ++) { ll temp; cin >> temp; bit.add(i, temp - last); // 在进行构造过程中，不能 bit.add(i, temp);，而应该用差分数组 last = temp; } for(ll i = 1; i <= q; i ++) { cin >> Q[i].id; if(Q[i].id == 1) cin >> Q[i].l >> Q[i].r >> Q[i].x; else if(Q[i].id == 2) cin >> Q[i].l; } for(ll i = 1; i <= q; i ++) { if(Q[i].id == 1) bit.inter_add(Q[i].l, Q[i].r, Q[i].x); else if(Q[i].id == 2) { ll res = bit.ask(Q[i].l); cout << res << endl; } } return 0; } - 1
- 2
- 3
- 4
- 5
- 6
- 7
- 8
- 9
- 10
- 11
- 12
- 13
- 14
- 15
- 16
- 17
- 18
- 19
- 20
- 21
- 22
- 23
- 24
- 25
- 26
- 27
- 28
- 29
- 30
- 31
- 32
- 33
- 34
- 35
- 36
- 37
- 38
- 39
- 40
- 41
- 42
- 43
- 44
- 45
- 46
- 47
- 48
- 49
- 50
- 51
- 52
- 53
- 54
- 55
- 56
- 57
- 58
- 59
- 60
- 61
- 62
- 63
- 64
- 65
- 66
- 67
- 68
- 69
- 70
- 71
- 72
- 73
- 74
- 75
- 76
- 77
- 78
- 79
- 80
- 81
- 82
/* time: 2022.9.9 区间修改，区间查询； */ #includeusing namespace std; typedef long long ll; const ll N = 1e6 + 3; struct BIT { ll t1[N], t2[N], n;//t1[N]是差分数组的树状数组，t2[N]是i*diff[i]的树状数组 //t1单点修改(这里是对差分数组单点修改) inline void add1(ll x, ll k) { for(; x <= n; x += x & -x) t1[x] += k; } //t2 单点修改（这里是对i*diffi单点修改） inline void add2(ll x, ll k) { for(; x <= n; x += x & -x) t2[x] += k; } //求 diff从1到x的前缀和 inline ll prefix1(ll x) { ll ans = 0; for(; x >= 1; x -= x & -x) ans += t1[x]; return ans; } //求i* diff的从1到x前缀和 inline ll prefix2(ll x) { ll ans = 0; for(; x >= 1; x -= x & -x) ans += t2[x]; return ans; } //区间查询，求原数组[l,r]的区间和 inline ll inter_sum(ll l, ll r) { return ((r+1)*prefix1(r) - prefix2(r)) - (l*prefix1(l - 1) - prefix2(l - 1)); } //区间修改，对元素组[l,r]+k inline void inter_add(ll l, ll r, ll k) { add1(l, k); add1(r + 1, -k); //同时需要对 i*diff这个数组进行修改 add2(l, l * k); add2(r + 1, - (r + 1) * k); } }; struct BIT bit; ll q; struct Node { ll id, l, r, x; }Q[N]; int main() { ios::sync_with_stdio(false); cin >> bit.n >> q; ll last = 0; for(ll i = 1;i <= bit.n; i ++) { ll temp; cin >> temp; bit.add1(i, temp - last); bit.add2(i, i * (temp - last)); last = temp; } for(ll i = 1; i <= q; i ++) { cin >> Q[i].id; if(Q[i].id == 1) cin >> Q[i].l >> Q[i].r >> Q[i].x; else if(Q[i].id == 2) cin >> Q[i].l >> Q[i].r; } for(ll i = 1; i <= q; i ++) { if(Q[i].id == 1) bit.inter_add(Q[i].l, Q[i].r, Q[i].x); else if(Q[i].id == 2) cout << bit.inter_sum(Q[i].l, Q[i].r) << endl; } return 0; } - 1
- 2
- 3
- 4
- 5
- 6
- 7
- 8
- 9
- 10
- 11
- 12
- 13
- 14
- 15
- 16
- 17
- 18
- 19
- 20
- 21
- 22
- 23
- 24
- 25
- 26
- 27
- 28
- 29
- 30
- 31
- 32
- 33
- 34
- 35
- 36
- 37
- 38
- 39
- 40
- 41
- 42
- 43
- 44
- 45
- 46
- 47
- 48
- 49
- 50
- 51
- 52
- 53
- 54
- 55
- 56
- 57
- 58
- 59
- 60
- 61
- 62
- 63
- 64
- 65
- 66
- 67
- 68
- 69
- 70
- 71
- 72
- 73
- 74
- 75
- 76
- 77
- 78
- 79
- 80
- 81
- 82
- 83
- 84
- 85
- 86
- 87
- 88
- 89
- 90
- 91
- 92
- 93
- 94
- 95
- 96
- 97
- 98
- 99
- 100
- 101
- 102
- 103
- 104
-
相关阅读:
Tungsten Fabric Rabbitmq故障处理
Apipost使用介绍
【ASP.NET小白零基础入门】从0部署ASP.NET开发环境，并成功运行一个汉服图片管理系统
LVS的实现NAT路由转发
题目0164-数组合并
Java_多态
python一键采集高质量陪玩，心动主播随心选......
c/c++ 静态代码检查工具
LeetCode算法心得——美丽塔 I（HashMap）
树莓派4B（64位）环境搭建
- 原文地址：https://blog.csdn.net/qq_45670253/article/details/126747640
