Codeforces Round #294 (Div. 2)
/*
水题
*/
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <string>
using namespace std; const int maxn = 1e6 + 10;
int a[maxn]; int main(void)
{
//freopen ("A.in", "r", stdin); string s1;
int suma = 0; int sumb = 0;
for (int i=1; i<=8; ++i)
{
cin >> s1;
for (int j=0; s1[j]!='\0'; ++j)
{
if (s1[j] == '.') continue;
else if (s1[j] == 'Q') suma += 9;
else if (s1[j] == 'R') suma += 5;
else if (s1[j] == 'B') suma += 3;
else if (s1[j] == 'N') suma += 3;
else if (s1[j] == 'P') suma += 1;
else if (s1[j] == 'q') sumb += 9;
else if (s1[j] == 'r') sumb += 5;
else if (s1[j] == 'b') sumb += 3;
else if (s1[j] == 'n') sumb += 3;
else if (s1[j] == 'p') sumb += 1;
}
} if (suma > sumb) cout << "White" << endl;
else if (suma < sumb) cout << "Black" << endl;
else cout << "Draw" << endl; return 0;
}
水 B. A and B and Compilation Errors
题意:三组数列,依次少一个,找出少了的两个数
思路:
1. 三次排序,逐个对比(如果没找到,那个数在上一个数列的末尾)
2. 求和做差,最简单!
#include <cstdio>
#include <algorithm>
#include <iostream>
using namespace std; const int maxn = 1e5 + 10;
int a[maxn];
int b[maxn];
int c[maxn]; int main(void)
{
//freopen ("B.in", "r", stdin); int n, x, y; while (~scanf ("%d", &n))
{
x = y = 0;
for (int i=1; i<=n; ++i)
{
scanf ("%d", &a[i]);
}
sort (a+1, a+1+n);
for (int i=1; i<=n-1; ++i)
{
scanf ("%d", &b[i]);
}
sort (b+1, b+1+n-1);
for (int i=1; i<=n-1; ++i)
{
if (a[i] == b[i]) continue;
else
{
x = a[i];
break;
}
}
if (x == 0) x = a[n];
for (int i=1; i<=n-2; ++i)
{
scanf ("%d", &c[i]);
}
sort (c+1, c+1+n-2);
for (int i=1; i<=n-2; ++i)
{
if (b[i] == c[i]) continue;
else
{
y = b[i];
break;
}
}
if (y == 0) y = b[n-1];
printf ("%d\n%d\n", x, y); } return 0;
} /*
#include <cstdio>
#include <algorithm>
#include <iostream>
using namespace std; const int maxn = 1e5 + 10;
int a[maxn];
int b[maxn];
int c[maxn];
int suma, sumb, sumc; int main(void)
{
//freopen ("B.in", "r", stdin); int n; while (~scanf ("%d", &n))
{
suma = sumb = sumc = 0;
for (int i=1; i<=n; ++i)
{
scanf ("%d", &a[i]); suma += a[i];
}
for (int i=1; i<=n-1; ++i)
{
scanf ("%d", &b[i]); sumb += b[i];
}
for (int i=1; i<=n-2; ++i)
{
scanf ("%d", &c[i]); sumc += c[i];
} printf ("%d\n%d\n", suma - sumb, sumb - sumc);
} return 0;
}
*/
构造 C. A and B and Team Training
题意:方案:高手1和菜鸟2 或者 高手2菜鸟1 三人组队求最大组队数
思路:
1. 高手加菜鸟每三个分开,在n,m的数字之内
2. 高手多,高手2;菜鸟多,菜鸟2 比较好理解
#include <cstdio>
#include <algorithm>
using namespace std; int main(void)
{
//freopen ("C.in", "r", stdin); int n, m; while (~scanf ("%d%d", &n, &m))
{
int ans = (n + m) / 3;
ans = min (ans, n);
ans = min (ans, m);
printf ("%d\n", ans);
} return 0;
} /*
#include <cstdio>
#include <algorithm>
using namespace std; int main(void)
{
//freopen ("C.in", "r", stdin); int n, m; while (~scanf ("%d%d", &n, &m))
{
int cnt = 0;
while (n && m && (n + m) >= 3)
{
if (n >= m)
{
n -= 2; m -= 1;
}
else
{
n -=1; m -= 2;
}
cnt++;
} printf ("%d\n", cnt); return 0;
}
*/
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