Uncowed Forces
2016-07-21 20:22
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Uncowed Forces
Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u
Submit Status
Description
Kevin Sun has just finished competing in Codeforces Round #334! The round was 120 minutes long and featured five problems with maximum point values of 500, 1000, 1500, 2000, and 2500, respectively. Despite the challenging tasks, Kevin was uncowed and bulldozed
through all of them, distinguishing himself from the herd as the best cowmputer scientist in all of Bovinia. Kevin knows his submission time for each problem, the number of wrong submissions that he made on each problem, and his total numbers of successful
and unsuccessful hacks. Because Codeforces scoring is complicated, Kevin wants you to write a program to compute his final score.
Codeforces scores are computed as follows: If the maximum point value of a problem is x, and Kevin submitted correctly at minute mbut
made w wrong submissions, then his score on that problem is
.
His total score is equal to the sum of his scores for each problem. In addition, Kevin's total score gets increased by 100 points for each successful hack, but gets decreased by 50 points
for each unsuccessful hack.
All arithmetic operations are performed with absolute precision and no rounding. It is guaranteed that Kevin's final score is an integer.
Input
The first line of the input contains five space-separated integers m1, m2, m3, m4, m5,
where mi (0 ≤ mi ≤ 119) is the time of Kevin's
last submission for problem i. His last submission is always correct and gets accepted.
The second line contains five space-separated integers w1, w2, w3, w4, w5,
where wi (0 ≤ wi ≤ 10) is Kevin's number of wrong
submissions on problem i.
The last line contains two space-separated integers hs and hu (0 ≤ hs, hu ≤ 20),
denoting the Kevin's numbers of successful and unsuccessful hacks, respectively.
Output
Print a single integer, the value of Kevin's final score.
Sample Input
Input
Output
Input
Output
Hint
In the second sample, Kevin takes 119 minutes on all of the problems. Therefore, he gets
of
the points on each problem. So his score from solving problems is
. Adding in 10·100 = 1000 points
from hacks, his total score becomes 3930 + 1000 = 4930.
题意:给定提交时间,和提交次数,hack成功和失败的次数,求分数
直接模拟(浮点数转整形时记得加0.5)
代码:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
double f(int x, int m, int w)
{
double xx = x;
double mm = m;
double ww = w;
return max(0.3*xx, (1- mm/250.0 )*xx - 50*ww) + 0.5;
}
int main()
{
int m1, m2, m3, m4, m5;
int w1, w2, w3, w4, w5;
while(~scanf("%d%d%d%d%d", &m1, &m2, &m3, &m4, &m5))
{
int ans = 0;
int hs, hu;
scanf("%d%d%d%d%d", &w1, &w2, &w3, &w4, &w5);
scanf("%d%d", &hs, &hu);
ans += f(500, m1, w1);
ans += f(1000, m2, w2);
ans += f(1500, m3, w3);
ans += f(2000, m4, w4);
ans += f(2500, m5, w5);
ans += 100*hs - 50*hu;
printf("%d\n", ans);
}
return 0;
}
Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u
Submit Status
Description
Kevin Sun has just finished competing in Codeforces Round #334! The round was 120 minutes long and featured five problems with maximum point values of 500, 1000, 1500, 2000, and 2500, respectively. Despite the challenging tasks, Kevin was uncowed and bulldozed
through all of them, distinguishing himself from the herd as the best cowmputer scientist in all of Bovinia. Kevin knows his submission time for each problem, the number of wrong submissions that he made on each problem, and his total numbers of successful
and unsuccessful hacks. Because Codeforces scoring is complicated, Kevin wants you to write a program to compute his final score.
Codeforces scores are computed as follows: If the maximum point value of a problem is x, and Kevin submitted correctly at minute mbut
made w wrong submissions, then his score on that problem is
.
His total score is equal to the sum of his scores for each problem. In addition, Kevin's total score gets increased by 100 points for each successful hack, but gets decreased by 50 points
for each unsuccessful hack.
All arithmetic operations are performed with absolute precision and no rounding. It is guaranteed that Kevin's final score is an integer.
Input
The first line of the input contains five space-separated integers m1, m2, m3, m4, m5,
where mi (0 ≤ mi ≤ 119) is the time of Kevin's
last submission for problem i. His last submission is always correct and gets accepted.
The second line contains five space-separated integers w1, w2, w3, w4, w5,
where wi (0 ≤ wi ≤ 10) is Kevin's number of wrong
submissions on problem i.
The last line contains two space-separated integers hs and hu (0 ≤ hs, hu ≤ 20),
denoting the Kevin's numbers of successful and unsuccessful hacks, respectively.
Output
Print a single integer, the value of Kevin's final score.
Sample Input
Input
20 40 60 80 100 0 1 2 3 4 1 0
Output
4900
Input
119 119 119 119 119 0 0 0 0 0 10 0
Output
4930
Hint
In the second sample, Kevin takes 119 minutes on all of the problems. Therefore, he gets
of
the points on each problem. So his score from solving problems is
. Adding in 10·100 = 1000 points
from hacks, his total score becomes 3930 + 1000 = 4930.
题意:给定提交时间,和提交次数,hack成功和失败的次数,求分数
直接模拟(浮点数转整形时记得加0.5)
代码:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
double f(int x, int m, int w)
{
double xx = x;
double mm = m;
double ww = w;
return max(0.3*xx, (1- mm/250.0 )*xx - 50*ww) + 0.5;
}
int main()
{
int m1, m2, m3, m4, m5;
int w1, w2, w3, w4, w5;
while(~scanf("%d%d%d%d%d", &m1, &m2, &m3, &m4, &m5))
{
int ans = 0;
int hs, hu;
scanf("%d%d%d%d%d", &w1, &w2, &w3, &w4, &w5);
scanf("%d%d", &hs, &hu);
ans += f(500, m1, w1);
ans += f(1000, m2, w2);
ans += f(1500, m3, w3);
ans += f(2000, m4, w4);
ans += f(2500, m5, w5);
ans += 100*hs - 50*hu;
printf("%d\n", ans);
}
return 0;
}
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