算法复习之前缀和【备战蓝桥杯】

一维前缀和

S[i] = a[1] + a[2] + ... a[i]
a[l] + ... + a[r] = S[r] - S[l - 1]
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二维前缀和

S[i, j] = 第i行j列格子左上部分所有元素的和
以(x1, y1)为左上角，(x2, y2)为右下角的子矩阵的和为：
S[x2, y2] - S[x1 - 1, y2] - S[x2, y1 - 1] + S[x1 - 1, y1 - 1]
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练习题

562. 壁画

#include 
#include 
#include 

using namespace std;

const int N = 5e6 + 10;
int n;
int s[N];
char str[N];

int main()
{
    int T;
    scanf("%d", &T);

    for (int x = 1; x <= T; x ++ )
    {
        scanf("%d", &n);
        scanf("%s", str + 1);

        memset(s, 0, sizeof s);
        for (int i = 1; i <= n; i ++ ) 
            s[i] += s[i - 1] + str[i] - '0';
        int k = (n - 1) / 2 + 1;

        int sum = 0;
        for (int i = k; i <= n; i ++ )
            sum = max(sum, s[i] - s[i - k]);

        printf("Case #%d: %d\n", x, sum);
    }

    return 0;
}
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795. 前缀和

#include 
#include 
#include 

using namespace std;

const int N = 1e5 + 10;
int n, m;
int s[N];

int main()
{
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i ++ ) 
    {
        scanf("%d", &s[i]);
        s[i] += s[i - 1];
    }

    while (m -- ) 
    {
        int l, r;
        scanf("%d%d", &l, &r);
        printf("%d\n", s[r] - s[l - 1]);
    }

    return 0;
}
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796. 子矩阵的和

#include 
#include 
#include 

using namespace std;

int n, m, q;
const int N = 1010;
int s[N][N];

int main()
{
    scanf("%d%d%d", &n, &m, &q);
    for (int i = 1; i <= n; i ++ ) 
        for (int j = 1; j <= m; j ++ ) 
            scanf("%d", &s[i][j]);
            
    for (int i = 1; i <= n; i ++ ) 
        for (int j = 1; j <= m; j ++ ) 
            s[i][j] += s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1];
            
    while(q -- )
    {
        int x1, y1, x2, y2;
        cin >> x1 >> y1 >> x2 >> y2;
        printf("%d\n", s[x2][y2] - s[x1 - 1][y2] - s[x2][y1 - 1] + s[x1 - 1][y1 - 1]);
    }
    
    return 0;
}
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1230. K倍区间

#include 
#include 
#include 
#include 

using namespace std;

typedef long long LL;
const int N = 100010;

int n, k;
LL s[N], cnt[N];

int main()
{
    scanf("%d%d", &n, &k);
    for (int i = 1; i <= n; i ++ ) 
    {
        scanf("%d", &s[i]);
        s[i] += s[i - 1];
    }
    
    // for (int i = 1; i <= n; i ++ ) printf("%d ", s[i]);
    
    LL res = 0;
    cnt[0] ++;
    for (int i = 1; i <= n; i ++ ) 
    {
        res += (LL)cnt[s[i] % k];
        cnt[s[i] % k] ++;
    }
    
    printf("%lld\n", res);
    
    return 0;
}
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4405. 统计子矩阵

#include 
#include 
#include 

using namespace std;

typedef long long LL;
const int N = 510;
int n, m, k;
int s[N][N];

int main()
{
    scanf("%d%d%d", &n, &m, &k);
    for (int i = 1; i <= n; i ++ ) 
        for (int j = 1; j <= m; j ++ ) 
        {
            scanf("%d", &s[i][j]);
            s[i][j] += s[i - 1][j];
        }
            
    LL res = 0;      
    // 枚举上下边界
    for (int i = 1; i <= n; i ++ ) 
        for (int j = i; j <= n; j ++ )
            // 双指针降低一层循环来枚举左右边界
            for (int l = 1, r = 1, sum = 0; r <= m; r ++ ) 
            {
                sum += s[j][r] - s[i - 1][r];
                while (sum > k) 
                {
                    sum -= s[j][l] - s[i - 1][l];
                    l ++;
                }
                res += r - l + 1;
            }
            
    printf("%lld\n", res);
    return 0;
}
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