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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 |
---|---|---|---|---|---|---|---|---|---|
#738289 | #9620. osu!mania | susanzhishen# | AC ✓ | 0ms | 3900kb | C++20 | 1019b | 2024-11-12 18:33:49 | 2024-11-12 18:33:50 |
Judging History
answer
#include <cstdio>
#include <iostream>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <cstring>
using namespace std;
typedef long long LL;
typedef long double LD;
void slv() {
LL ppmax, a, b, c, d, e, f;
cin >> ppmax >> a >> b >> c >> d >> e >> f;
LD ans1 = (LD)(300 * a + 300 * b + 200 * c + 100 * d + 50 * e + 0 * f) / (3 * (a + b + c + d + e + f));
LL sum = a + b + c + d + e + f;
LL v = 320 * a + 300 * b + 200 * c + 100 * d + 50 * e + 0 * f - 256 * sum;
LL ans2;
if(v < 0) ans2 = 0;
else ans2 = roundl((LD)(v * 5 * ppmax) / (320 * sum));
// LD ans2 = max(0.0l, ((LD)(320 * a + 300 * b + 200 * c + 100 * d + 50 * e + 0 * f) / (320 * (a + b + c + d + e + f))) - 0.8) * 5 * ppmax;
printf("%.2Lf%% %lld\n", ans1, ans2);
// printf("%.100Lf\n", ans2);
}
int main() {
int t;
cin >> t;
while(t--) {
slv();
}
return 0;
}
/*
2
630
3029 2336 377 41 10 61
3000
20000 10000 0 0 0 0
*/
详细
Test #1:
score: 100
Accepted
time: 0ms
memory: 3900kb
input:
18 1279 4624 4458 1109 220 103 314 753 3604 3204 391 33 9 29 807 5173 3986 763 84 29 96 718 576 461 60 5 2 7 947 4058 3268 764 169 42 158 568 2660 1731 161 16 6 15 641 4181 3126 656 56 10 43 630 3029 2336 377 41 10 61 529 1991 1354 181 11 9 5 1802 8321 2335 115 19 11 27 1645 3965 1087 41 6 1 13 1688...
output:
91.54% 543 97.40% 543 95.75% 523 97.12% 513 93.38% 499 98.16% 444 96.19% 430 96.20% 423 97.74% 400 99.19% 1604 99.38% 1482 99.14% 1465 98.53% 1251 100.00% 2688 100.00% 1792 100.00% 3000 52.78% 0 0.00% 0
result:
ok 18 lines