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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
#768698#7785. Three RectanglesSGColin#WA 1ms4032kbC++144.2kb2024-11-21 13:45:572024-11-21 13:45:58

Judging History

你现在查看的是最新测评结果

  • [2024-11-21 13:45:58]
  • 评测
  • 测评结果:WA
  • 用时:1ms
  • 内存:4032kb
  • [2024-11-21 13:45:57]
  • 提交

answer

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef tuple<int, int, int> tii;
typedef vector<int> vi;

inline int rd() {
    int x = 0;
    bool f = 0;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) f |= (c == '-');
    for (; isdigit(c); c = getchar()) x = x * 10 + (c ^ 48);
    return f ? -x : x;
}

#define eb emplace_back
#define all(s) (s).begin(), (s).end()
#define rep(i, a, b) for (int i = (a); i <= (b); ++i)
#define per(i, a, b) for (int i = (a); i >= (b); --i)

constexpr int N = 5007;

struct rectan {
    int h, w;
    bool operator < (const rectan &obj) const {return h == obj.h ? (w < obj.w) : (h < obj.h);}
    bool operator > (const rectan &obj) const {return obj < (*this);}
    bool operator == (const rectan &obj) const {return !((*this) < obj || (*this) > obj);}
    bool operator != (const rectan &obj) const {return !(*this == obj);}
    bool operator <= (const rectan &obj) const {return !(*this > obj);}
    bool operator >= (const rectan &obj) const {return !(*this < obj);}
};

const int mod = 1000000007;

inline void calc() {
    rectan R{rd(), rd()};
    vector<rectan> r;
    int HW = 0, H = 0, W = 0;
    rep(i, 1, 3) {
        rectan rr{rd(), rd()};
        if(rr.h == R.h && rr.w == R.w) HW++;
        if(rr.h == R.h) H++;
        if(rr.w == R.w) W++;
        r.eb(rr);
    }
    if (HW > 0) {
        int ans = 1;
        for(auto rr : r) {
            int res = 1ll * (R.h - rr.h + 1) * (R.w - rr.w + 1) % mod;
            ans = (1ll * ans * res) % mod;
        }
        printf("%d\n", ans);
        return;
    }
    if (W > 0) {
        int ans = 0;
        if (W == 3) {
            sort(all(r));
            do {
                if (r[0].h + r[2].h >= R.h) ans = (ans + R.h - 1 - r[1].h) % mod;
                else {
                    int low = max(R.h - r[2].h - r[1].h + 1, 2);
                    int up = min(R.h - r[1].h, r[0].h + 1);
                    if (low <= up) ans = (ans + up - low + 1) % mod;
                }
            }while (next_permutation(all(r)));
            sort(all(r));
            if (r[0].h + r[2].h >= R.h) ans = (ans + 2) % mod;
            if (r[1].h + r[2].h >= R.h) ans = (ans + 4) % mod;
        }
        else if (W == 2) {
            if (r[0].w != R.w) swap(r[0], r[2]);
            if (r[1].w != R.w) swap(r[1], r[2]);
            if (r[0].h + r[1].h >= R.h) ans = (ans + 2ll * (R.h - r[2].h +1) * (R.w - r[2].w + 1) % mod) % mod;
        }
        else if (W == 1) {
            if (r[1].w == R.w) swap(r[1], r[0]);
            if (r[2].w == R.w) swap(r[2], r[0]);
            if(r[1].w + r[2].w >= R.w && r[1].h + r[0].h >= R.h && r[2].h + r[0].h >= R.h) ans = (ans + 4) % mod;
        }
        printf("%d\n", ans);
        return;
    }
    if (H > 0) {
        int ans = 0;
        if (H == 3) {
            sort(all(r));
            do {
                if (r[0].w + r[2].w >= R.w) ans = (ans + R.w - r[1].w - 1) % mod;
                else {
                    int low = max(R.w - r[2].w - r[1].w + 1, 2);
                    int up = min(R.w - r[1].w, r[0].w + 1);
                    if (low <= up) ans = (ans + up - low + 1) % mod;
                }
            }while (next_permutation(all(r)));
            sort(all(r));
            if (r[0].w + r[2].w >= R.w) ans = (ans + 2) % mod;
            if (r[1].w + r[2].w >= R.w) ans = (ans + 4) % mod;
        }
        else if (H == 2) {
            if (r[0].h != R.h) swap(r[0], r[2]);
            if (r[1].h != R.h) swap(r[1], r[2]);
            if (r[0].w + r[1].w >= R.w) ans = (ans + 2ll * (R.w - r[2].w +1) * (R.h - r[2].h + 1) % mod) % mod;
        }
        else if (H == 1) {
            if (r[1].h == R.h) swap(r[1], r[0]);
            if (r[2].h == R.h) swap(r[2], r[0]);
            if(r[1].h + r[2].h >= R.h && r[1].w + r[0].w >= R.w && r[2].w + r[0].w >= R.w) ans = (ans + 4) % mod;
        }
        printf("%d\n", ans);
        return;
    }
    puts("0");
    return;
}

int main() {
    per(i, rd(), 1) calc();
    return 0;
}

/*
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
*/

详细

Test #1:

score: 100
Accepted
time: 0ms
memory: 4032kb

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: 3808kb

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: 4032kb

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: 3744kb

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
3
888888883
456790164
0
0
0
0
0
6
222222208
555555531
135802502
0
0
0
0
6
222222208
222222208
333333309
814814847
0
0
0
3
222222208
222222208
222222208
111111087
493827185
0
0
6
222222208
111111106
222222208
222222208
888888872
172839523
0
6
222222208
222222208
222222208
111111106
222222...

result:

wrong answer 7th numbers differ - expected: '6', found: '3'