一.前缀和(一维,二维,三维)
模板
/*
一维前缀和
s[i] = s[i-1] + a[i];
查询 l - r 的和
s[r] - s[l-1]
*/
/*
二维前缀和
s[i][j] = s[i-1][j] + s[i][j-1] - s[i-1][j-1] + a[i][j];
查询 (x1,y1) - (x2,y2) 的和
s[x2][y2] - s[x1 - 1][y2] - s[x2][y1 - 1] + s[x1 - 1][y1 - 1]
*/
/*
三维前缀和
s[i][j][k] = s[i-1][j][k] + s[i][j-1][k] + s[i][j][k-1]
- s[i-1][j-1][k] - s[i-1][j][k-1] - s[i][j-1][k-1]
+ s[i-1][j-1][k-1] + a[i][j][k];
查询(x1,y1,z1) - (x2,y2,z2) 的和
s[x2][y2][z2] - s[x1-1][y1-1][z1-1] - s[x1-1][y2][z2] - s[x2][y1-1][z2] - s[x2][y2][z1-1]
+ s[x1-1][y1-1][z2] + s[x1-1][y2][z1-1] + s[x2][y1-1][z1-1];
*/
例题
Kobolds and Catacombs
思路:能够分开的点 前面的所有数一定小于后面所有数 可以利用前缀和来判断 s表示原数组的前缀和 presum 表示 排成升序的前缀和 当s[i] == presum[i] 时 ans++
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N = 1000010;
LL a[N];
LL presum[N];
LL s[N];
int ans;
int main()
{
int n;
cin >> n;
for(int i = 1;i <= n;i++)
{
cin >> a[i];
s[i] += s[i-1] + (LL)a[i];
}
sort(a + 1,a + n + 1);
for(int i = 1;i <= n;i++) presum[i] += presum[i-1] + (LL)a[i];
bool flag = true;
for(int i = 1;i <= n;i++)
{
if(s[i] == presum[i]) ans++;
else flag = false;
}
cout << ans << endl;
}
Cuboid Sum Query(三维前缀和 模板)
#include<bits/stdc++.h>
using namespace std;
const int N = 110;
typedef long long LL;
int a[N][N][N];
LL s[N][N][N];
int main()
{
int n;
cin >> n;
for(int i = 1;i <= n;i++)
for(int j = 1;j <= n;j++)
for(int k = 1;k <= n;k++)
cin >> a[i][j][k];
for(int i = 1;i <= n;i++)
for(int j = 1;j <= n;j++)
for(int k = 1;k <= n;k++)
s[i][j][k] = s[i-1][j][k] + s[i][j-1][k] + s[i][j][k-1]
- s[i-1][j-1][k] - s[i-1][j][k-1] - s[i][j-1][k-1]
+ s[i-1][j-1][k-1] + (LL)a[i][j][k];
int Q;
cin >> Q;
while(Q--)
{
int x1,x2,y1,y2,z1,z2;
cin >> x1 >> x2 >> y1 >> y2 >> z1 >> z2;
LL sum = s[x2][y2][z2] - s[x1-1][y1-1][z1-1]
- s[x1-1][y2][z2] - s[x2][y1-1][z2] - s[x2][y2][z1-1]
+ s[x1-1][y1-1][z2] + s[x1-1][y2][z1-1] + s[x2][y1-1][z1-1] ;
cout << sum