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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 |
---|---|---|---|---|---|---|---|---|---|
#771378 | #5444. Tavern Chess | ucup-team3702# | RE | 22ms | 11112kb | C++14 | 2.6kb | 2024-11-22 12:19:36 | 2024-11-22 12:19:37 |
Judging History
answer
#include <bits/stdc++.h>
using namespace std;
#define rep(i, d, u) for(int i = d; i <= u; ++i)
const int _maxn = 7, _zinf = 1000000001;
int n, m, a[_maxn], b[_maxn];
struct node {
int le[_maxn], ri[_maxn];
long long lt[_maxn], rt[_maxn];
node atak(int la, int ra, bool st) {
if(st) {
rt[ra] += a[la];
} else {
lt[la] += b[ra];
}
node re = *this;
if((re . le[la] -= b[ra]) < 0) {
re . le[la] = 0;
}
if((re . ri[ra] -= a[la]) < 0) {
re . ri[ra] = 0;
}
return re;
}
int chek() {
int la = 0, ra = 0;
rep(i, 0, n - 1) {
if(le[i]) {
la = 1;
break;
}
}
rep(i, 0, m - 1) {
if(ri[i]) {
ra = 1;
break;
}
}
return la << 1 | ra;
}
} orig;
bool operator<(node le, node ri) {
rep(i, 0, n - 1) {
if(le . le[i] != ri . le[i]) {
return le . le[i] < ri . le[i];
}
}
rep(i, 0, m - 1) {
if(le . ri[i] != ri . ri[i]) {
return le . ri[i] < ri . ri[i];
}
}
rep(i, 0, n - 1) {
if(le . lt[i] != ri . lt[i]) {
return le . lt[i] < ri . lt[i];
}
}
rep(i, 0, m - 1) {
if(le . rt[i] != ri . rt[i]) {
return le . rt[i] < ri . rt[i];
}
}
return 0;
}
struct dats {
double le, ri, bl;
dats() {}
dats(double fi, double se, double th) : le(fi), ri(se), bl(th) {}
};
dats operator+(dats le, dats ri) {
return dats(le . le + ri . le, le . ri + ri . ri, le . bl + ri . bl);
}
dats operator/(dats le, int ri) {
return dats(le . le / ri, le . ri / ri, le . bl / ri);
}
map<node, dats> maps[2];
dats dfs1(bool at, node st) {
switch(st . chek()) {
case 0 : return dats(0, 0, 1);
case 1 : return dats(0, 1, 0);
case 2 : return dats(1, 0, 0);
}
if(maps[at] . count(st)) {
return maps[at][st];
}
int mi = _zinf, po, cn = 0;
dats re(0, 0, 0);
if(at) {
rep(i, 0, m - 1) {
if(st . ri[i] && mi > st . rt[i]) {
mi = st . rt[i], po = i;
}
}
rep(i, 0, n - 1) {
if(st . le[i]) {
re = re + dfs1(0, st . atak(i, po, at)), ++cn;
}
}
} else {
rep(i, 0, n - 1) {
if(st . le[i] && mi > st . lt[i]) {
mi = st . lt[i], po = i;
}
}
rep(i, 0, m - 1) {
if(st . ri[i]) {
re = re + dfs1(1, st . atak(po, i, at)), ++cn;
}
}
}
return maps[at][st] = re / cn;
}
int main() {
cin >> n >> m;
rep(i, 0, n - 1) {
cin >> a[i], orig . le[i] = a[i];
}
rep(i, 0, m - 1) {
cin >> b[i], orig . ri[i] = b[i];
}
dats rans;
if(n > m) {
rans = dfs1(0, orig);
} else if(n < m) {
rans = dfs1(1, orig);
} else {
rans = (dfs1(0, orig) + dfs1(1, orig)) / 2;
}
printf("%.15lf\n%.15lf\n%.15lf\n", rans . le, rans . ri, rans . bl);
return 0;
}
詳細信息
Test #1:
score: 100
Accepted
time: 0ms
memory: 3856kb
input:
2 3 2 5 3 4 1
output:
0.125000000000000 0.750000000000000 0.125000000000000
result:
ok 3 numbers
Test #2:
score: 0
Accepted
time: 22ms
memory: 11112kb
input:
6 6 1 1 4 5 1 4 1 1 4 5 1 4
output:
0.241867283950617 0.241867283950617 0.516265432098766
result:
ok 3 numbers
Test #3:
score: 0
Accepted
time: 9ms
memory: 7896kb
input:
7 7 1 1 1 1 1 1 1 1 1 1 1 1 1 1
output:
0.000000000000000 0.000000000000000 1.000000000000000
result:
ok 3 numbers
Test #4:
score: 0
Accepted
time: 0ms
memory: 3808kb
input:
1 7 7 1 1 1 1 1 1 1
output:
0.000000000000000 0.000000000000000 1.000000000000000
result:
ok 3 numbers
Test #5:
score: 0
Accepted
time: 0ms
memory: 3968kb
input:
2 3 736618938 652769331 328875880 97571721 44608905
output:
1.000000000000000 0.000000000000000 0.000000000000000
result:
ok 3 numbers
Test #6:
score: -100
Runtime Error
input:
5 4 53585130 731696211 668322278 611205195 158818781 569587984 776042583 745745433 330119007