QOJ.ac

QOJ

ID题目提交者结果用时内存语言文件大小提交时间测评时间
#602318#7785. Three RectanglesAlphaZeWA 1ms3848kbC++147.7kb2024-09-30 22:48:062024-09-30 22:49:12

Judging History

你现在查看的是最新测评结果

  • [2024-09-30 22:49:12]
  • 评测
  • 测评结果:WA
  • 用时:1ms
  • 内存:3848kb
  • [2024-09-30 22:48:06]
  • 提交

answer

#include <bits/stdc++.h>

using namespace std;

inline int read() {
    int x = 0, f = 0; char ch = getchar();
    while (!isdigit(ch)) f = ch == '-', ch = getchar();
    while (isdigit(ch)) x = (x << 3) + (x << 1) + (ch ^ 48), ch = getchar();
    return f ? -x : x;
}   

const int mod = 1e9 + 7;
void MulMod(int &p, int k) { p = 1ll * p * k % mod; }
void AddMod(int &p, int k) { p = (p + k) % mod; }
int ans;
int A, B;
int a[3], b[3], s[3], p[3];
bool can[3][2];

bool chk() {
    bool fl = 0;
    for (int i = 0; i < 3; ++i)
        fl |= a[i] == A && b[i] == B;
    if (fl) {
        ans = 1;
        for (int i = 0; i < 3; ++i)
            MulMod(ans, 1ll * (A - a[i] + 1) * (B - b[i] + 1) % mod);
    }
    return fl;
}

int chk2(int step, int U, int D, int L, int R) {
    // printf("chk2(%d, %d, %d, %d, %d); \n", step, U, D, L, R);
    if (step == 3) {
        if (D - U <= 1 || R - L <= 1) return 1;
        return 0;
    }
    if (s[step] == 1) return chk2(step + 1, max(U, a[step]), D, L, R); 
    else if (s[step] == 3) return chk2(step + 1, U, min(D, A + 1 - a[step]), L, R);
    else if (s[step] == 2) return chk2(step + 1, U, D, max(L, b[step]), R);
    else return chk2(step + 1, U, D, L, min(R, B + 1 - b[step]));
}

int calc2(int step, int U, int D, int L, int R, bool fl0, bool fl1) {
    // printf("calc2(%d, %d, %d, %d, %d, %d, %d); \n", step, U, D, L, R, fl0, fl1);
    if (step == 2) {
        if (fl0 && fl1) {
            int aa = D - U - 1, bb = R - L - 1;
            if (aa <= a[p[step]] && bb <= b[p[step]]) return 1;
            return 0;
        } else if (fl1) {
           if (D - U <= 1) return 1ll * (A - a[p[step]] + 1) * (B - b[p[step]] + 1) % mod;
            else {
                if (b[p[step]] != B) return 0;
                else {
                    int aa = D - U - 1;
                    if (a[p[step]] < aa) return 0;
                    int res = a[p[step]] - aa + 1;
                    res = min(res, U + 1);
                    res = min(res, A + 1 - D + 1);
                    return res;
                }
            }
        } else {
            if (R - L <= 1) return 1ll * (A - a[p[step]] + 1) * (B - b[p[step]] + 1) % mod;
            else {
                if (a[p[step]] != A) return 0;
                else {
                    int bb = R - L - 1; 
                    if (b[p[step]] < bb) return 0;
                    int res = b[p[step]] - bb + 1;
                    res = min(res, L + 1);
                    res = min(res, B + 1 - R + 1);
                    return res;
                }
            }           
        }
    }
    if (s[p[step]] == 1) return calc2(step + 1, max(U, a[p[step]]), D, L, R, fl0, 1); 
    else if (s[p[step]] == 3) return calc2(step + 1, U, min(D, A + 1 - a[p[step]]), L, R, fl0, 1);
    else if (s[p[step]] == 2) return calc2(step + 1, U, D, max(L, b[p[step]]), R, 1, fl1);
    else return calc2(step + 1, U, D, L, min(R, B + 1 - b[p[step]]), 1, fl1);
}

int calc3() {
    // puts("calc3()");
    // for (int i = 0; i < 3; ++i) printf("%d ", p[i]); putchar('\n');
    int res = 0;
    if (s[p[0]] == 1 || s[p[0]] == 3) {
        int aa = A - a[p[0]];
        bool fl3 = b[p[1]] == B, fl4 = b[p[2]] == B;
        bool fl1 = a[p[1]] >= aa && b[p[1]] == B, fl2 = a[p[2]] >= aa && b[p[2]] == B;
        if (fl1 && fl2) {
            AddMod(res, 1ll * (A - a[p[2]] + 1) * (B - b[p[2]] + 1) % mod);
            AddMod(res, 1ll * (A - a[p[1]] + 1) * (B - b[p[1]] + 1) % mod);
            AddMod(res, mod - 1);
        } else if (fl1) {
            AddMod(res, 1ll * (A - a[p[2]] + 1) * (B - b[p[2]] + 1) % mod);
            if (fl3 && fl4) AddMod(res, mod - 1);
        } else if (fl2) {
            AddMod(res, 1ll * (A - a[p[1]] + 1) * (B - b[p[1]] + 1) % mod);
            if (fl3 && fl4) AddMod(res, mod - 1);
        } 
        if (fl3 && fl4) {
            s[p[1]] = s[p[0]]; 
            if (s[p[1]] > 2) s[p[1]] -= 2; else s[p[1]] += 2;
            if (!fl1) AddMod(res, calc2(0, 0, A + 1, 0, B + 1, 0, 0));
            swap(p[1], p[2]);
            s[p[1]] = s[p[0]]; 
            if (s[p[1]] > 2) s[p[1]] -= 2; else s[p[1]] += 2;
            if (!fl2) AddMod(res, calc2(0, 0, A + 1, 0, B + 1, 0, 0));
            swap(p[1], p[2]);
        } else {
            if (fl1 || fl2);
            else if (a[p[1]] < aa || a[p[2]] < aa);
            else if (b[p[1]] + b[p[2]] < B);
            else AddMod(res, 2);
        }

    } else {
        int bb = B - b[p[0]];
        bool fl3 = a[p[1]] == A, fl4 = a[p[2]] == A;
        bool fl1 = b[p[1]] >= bb && a[p[1]] == A, fl2 = b[p[2]] >= bb && a[p[2]] == A;
        if (fl1 && fl2) {
            AddMod(res, 1ll * (A - a[p[2]] + 1) * (B - b[p[2]] + 1) % mod);
            AddMod(res, 1ll * (A - a[p[1]] + 1) * (B - b[p[1]] + 1) % mod);
            AddMod(res, mod - 1);
        } else if (fl1) {
            AddMod(res, 1ll * (A - a[p[2]] + 1) * (B - b[p[2]] + 1) % mod);
            if (fl3 && fl4) AddMod(res, mod - 1);
        } else if (fl2) {
            AddMod(res, 1ll * (A - a[p[1]] + 1) * (B - b[p[1]] + 1) % mod);
            if (fl3 && fl4) AddMod(res, mod - 1);
        } 
        if (fl3 && fl4) {
            s[p[1]] = s[p[0]]; 
            if (s[p[1]] > 2) s[p[1]] -= 2; else s[p[1]] += 2;
            if (!fl1) AddMod(res, calc2(0, 0, A + 1, 0, B + 1, 0, 0));
            swap(p[1], p[2]);
            s[p[1]] = s[p[0]]; 
            if (s[p[1]] > 2) s[p[1]] -= 2; else s[p[1]] += 2;
            if (!fl2) AddMod(res, calc2(0, 0, A + 1, 0, B + 1, 0, 0));
            swap(p[1], p[2]);
        } else {
            if (fl1 || fl2);
            else if (b[p[1]] < bb || b[p[2]] < bb);
            else if (a[p[1]] + a[p[2]] < A);
            else AddMod(res, 2);
        }       
    }
    return res;
}

int calc() {
    // for (int i = 0; i < 3; ++i) printf("%d ", s[i]); putchar('\n');
    int cnt = 0, res = 0;
    for (int i = 0; i < 3; ++i) cnt += s[i] != 0;
    if (!cnt) return 0;
    else if (cnt == 1) {
        int nm = 0;
        for (int i = 0; i < 3; ++i) {
            if (!s[i]) p[++nm] = i;
            else p[0] = i;
        }
        return calc3();        
    } else if (cnt == 2) {
        int nm = -1;
        for (int i = 0; i < 3; ++i) {
            if (s[i]) p[++nm] = i;
            else p[2] = i;
        }
        // printf("->"); for (int i = 0; i < 3; ++i) printf("%d ", p[i]); putchar('\n');
        return mod - calc2(0, 0, A + 1, 0, B + 1, 0, 0);
        
    } else {
        return chk2(0, 0, A + 1, 0, B + 1);
    }
}

void work() {
    A = read(); B = read();
    for (int i = 0; i < 3; ++i) {
        a[i] = read(), b[i] = read();
        can[i][0] = can[i][1] = 0;
    }
    ans = 0;
    if (chk()) {
        printf("%d\n", ans); 
    } else {       
        for (int i = 0; i < 3; ++i) {
            can[i][1] = b[i] == B; // 上下放
            can[i][0] = a[i] == A; // 左右放
        }
        for (int i = 0; i <= 4; ++i) {
            if ((!i) || can[0][i & 1]) {
                for (int j = 0; j <= 4; ++j) {
                    if ((!j) || can[1][j & 1]) {
                       for (int k = 0; k <= 4; ++k) {
                           if ((!k) || can[2][k & 1]) {
                                s[0] = i; s[1] = j; s[2] = k;
                                AddMod(ans, calc());
                                // printf("ans = %d; \n", ans);
                           }
                       }                   
                    }
                }                
            }
        }
        printf("%d\n", ans);
    }
}

int main() {
    // freopen("I.out", "w", stdout);
    int Tt = read();
    while (Tt--) work();
    return 0;
}

詳細信息

Test #1:

score: 100
Accepted
time: 0ms
memory: 3832kb

input:

5
2 2
1 1
1 1
1 1
2 2
1 1
1 2
1 2
2 2
1 1
1 2
2 1
2 2
1 2
1 2
1 2
2 2
1 2
1 2
2 1

output:

0
8
4
6
4

result:

ok 5 number(s): "0 8 4 6 4"

Test #2:

score: 0
Accepted
time: 0ms
memory: 3816kb

input:

4
1 3
1 1
1 2
1 3
1 4
1 1
1 2
1 3
1 5
1 1
1 2
1 3
1 6
1 1
1 2
1 3

output:

6
12
14
6

result:

ok 4 number(s): "6 12 14 6"

Test #3:

score: 0
Accepted
time: 0ms
memory: 3848kb

input:

1
1000000000 1000000000
1 1
1 1
1000000000 1000000000

output:

2401

result:

ok 1 number(s): "2401"

Test #4:

score: -100
Wrong Answer
time: 1ms
memory: 3844kb

input:

729
999999999 111111111
111111111 111111111
111111111 111111111
111111111 111111111
999999999 111111111
111111111 111111111
222222222 111111111
111111111 111111111
999999999 111111111
111111111 111111111
111111111 111111111
333333333 111111111
999999999 111111111
111111111 111111111
444444444 111111...

output:

0
0
0
0
0
0
6
777777753
456790164
0
0
0
0
0
6
222222208
555555531
135802502
0
0
0
0
6
222222208
222222208
333333309
814814847
0
0
0
6
222222208
222222208
222222208
111111087
493827185
0
0
6
222222208
222222208
222222208
222222208
888888872
172839523
0
6
222222208
222222208
222222208
222222208
222222...

result:

wrong answer 98th numbers differ - expected: '333333309', found: '555555531'