[GXOI/GZOI2019]与或和 题解

开始完全没思路

在洛谷看到样例一，突发奇想，决定先做一下元素只有0/1的情况

发现子任务1是全1子矩阵

        子任务2是总子矩阵个数减去全0子矩阵

发现全0/1矩阵可以构造单调栈解决。具体做法：前缀和求出每个格子上面有多少颜色为0/1的格子(是0是1有求子任务1/2决定)，然后发现可以每次在单调栈中找出相邻的两个值，算出内部区块的面积，多次累加后发现刚好是全0/1子矩阵的个数

小技巧：把单调队列的第0项的坐标置0，可以避免特判

让后求总子矩阵个数也很简单，递推解决(我数学不好，瑟瑟发抖)

公式：　　ff[i][j] = ff[i - 1][j] + ff[i][j - 1] - ff[i - 1][j - 1] + i * j;

那么总子矩阵个数即为f[n][n]

让后向元素任意值得矩阵迈进

发现恰好可以以二进制来展开获得0/1矩阵

代码：

for (register int i = 0; i < n; ++i)
    for (register int j = 0; j < n; ++j)
        a[i][j] = (atot[i][j] & (1 << flr)) ? 1 : 0;

atot为读入数组，a为需要的0/1数组

flr表示现在在二进制的flr位

让后算出来的答案乘以(1 << flr)累加到总答案上

打完以后发现30分

快速浏览代码没找到错误(我太菜了)

后来点开了题解，正准备浏jie览jian，突然发现有一处没有MOD，

MOD了以后果断(???)AC...

贴个代码

#include <cstdio> 
#define ll long long

const ll MOD = 1e9+7;

inline ll read(){
    ll x = 0; int zf = 1; char ch = ' ';
    while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
    if (ch == '-') zf = -1, ch = getchar();
    while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar(); return x * zf;
}

ll atot[1005][1005];
int a[1005][1005];
int sum[1005][1005];

ll ff[1005][1005];

ll or_init;

struct Node{
  ll pos;
  ll hei;
} ddstk[1005];
int top;


int main(){
    int n = read();
    for (register int i = 0; i < n; ++i)
        for (register int j = 0; j < n; ++j)
            atot[i][j] = read();
    for (register int i = 1; i <= n; ++i){
        for (register int j = 1; j <= n; ++j){
            ff[i][j] = ff[i - 1][j] + ff[i][j - 1] - ff[i - 1][j - 1] + i * j;
            if (ff[i][j] < 0)
                ff[i][j] += MOD;
            ff[i][j] %= MOD;
        }
    }
    or_init = ff[n][n];
    ll ans1 = 0, cur_ans1, ans2 = 0, cur_ans2;
    ddstk[0].pos = 0;
    for (int flr = 0; flr < 31; ++flr){
        for (register int i = 0; i < n; ++i)
            for (register int j = 0; j < n; ++j)
                a[i][j] = (atot[i][j] & (1 << flr)) ? 1 : 0;
        //getAnd
        for (register int i = 0; i < n; ++i)
            for (register int j = 0; j < n; ++j)
                if (i != 0)
                    sum[i][j] = (a[i][j] == 1) ? sum[i - 1][j] + 1 : 0;
                else
                    sum[i][j] = (a[i][j] == 1) ? 1 : 0;
        cur_ans1 = 0;
        for (register int i = 0; i < n; ++i){
            top = 0;
            for (register int k = 0; k < n; ++k){
                while (top){
                    if (sum[i][k] <= ddstk[top].hei)
                        --top;
                    else
                        break;
                }
                ddstk[++top].pos = k + 1;
                ddstk[top].hei = sum[i][k];
                for (int l = top; l >= 1; --l){
                    cur_ans1 += ddstk[l].hei * (ddstk[l].pos - ddstk[l - 1].pos);
                    cur_ans1 %= MOD;
                }
            }
        }
        ans1 += (cur_ans1 * ((1ll << flr) % MOD)) % MOD;
        ans1 %= MOD;
        //getOr
        for (register int i = 0; i < n; ++i)
            for (register int j = 0; j < n; ++j)
                if (i != 0)
                    sum[i][j] = (a[i][j] == 0) ? sum[i - 1][j] + 1 : 0;
                else
                    sum[i][j] = (a[i][j] == 0) ? 1 : 0;
        cur_ans2 = 0;
        for (register int i = 0; i < n; ++i){
            top = 0;
            for (register int k = 0; k < n; ++k){
                while (top){
                    if (sum[i][k] <= ddstk[top].hei)
                        --top;
                    else
                        break;
                }
                ddstk[++top].pos = k + 1;
                ddstk[top].hei = sum[i][k];
                for (int l = top; l >= 1; --l){
                    cur_ans2 += ddstk[l].hei * (ddstk[l].pos - ddstk[l - 1].pos);
                    cur_ans2 %= MOD;
                }
            }
        }
        cur_ans2 = (or_init - cur_ans2 + MOD) % MOD;
        ans2 += (cur_ans2 * ((1ll << flr) % MOD)) % MOD;
        ans2 %= MOD;
    }
    printf("%lld %lld", ans1, ans2);
    return 0;
}