A. Decision Making
A题就是一句话的题目，比较str[len/2]和str[len/2-1]即可，不同NOT,反之DO
代码：
#include <iostream>
#include <cstdio>
#include <string>
#include <string.h>
#include <map>
#include <vector>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <queue>
#include <set>
#include <stack>
#include <functional>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <cassert>
#include <bitset>
#include <stack>
#include <ctime>
#include <list>
#define INF 0x7fffffff
#define max3(a,b,c) (max(a,b)>c?max(a,b):c)
#define min3(a,b,c) (min(a,b)<c?min(a,b):c)
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
int QuickMod(int a,int b,int n)
{
int r = 1;
while(b)
{
if(b&1)
r = (r*a)%n;
a = (a*a)%n;
b >>= 1;
}
return r;
}
int main()
{
string str;
int n;
scanf("%d", &n);
while(n--)
{
cin >> str;
int len = str.length();
if(str[len/2] != str[len/2 - 1])
printf("Do-it-Not\n");
else
printf("Do-it\n");
}
return 0;
}
B. Log Books
这道题目我们做了好久，谔谔，题意出了问题。刚刚我又重新敲了一遍。就是给你白天和晚上飞行的时间。问你是否符合要求即可。就是判断区间的相交情况。按照night > sun的变换，统计飞行时间即可。
代码：
#include <iostream>
#include <cstdio>
#include <string>
#include <string.h>
#include <map>
#include <vector>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <queue>
#include <set>
#include <stack>
#include <functional>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <cassert>
#include <bitset>
#include <stack>
#include <ctime>
#include <list>
#define INF 0x7fffffff
#define max3(a,b,c) (max(a,b)>c?max(a,b):c)
#define min3(a,b,c) (min(a,b)<c?min(a,b):c)
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
int QuickMod(int a,int b,int n)
{
int r = 1;
while(b)
{
if(b&1)
r = (r*a)%n;
a = (a*a)%n;
b >>= 1;
}
return r;
}
struct node
{
int h;
int m;
};
int cal(int x, int y)
{
return x*60 + y;
}
int limit1 = 3000;
int limit2 = 600;
int main()
{
int n;
while(scanf("%d", &n) && n)
{
node sunup, sundown, flys, flye;
int flag = 0;
int SUMS = 0;
for(int i = 0; i < n; ++i)
{
scanf("%d:%d%d:%d%d:%d%d:%d", &sunup.h, &sunup.m, &sundown.h, &sundown.m,
&flys.h, &flys.m, &flye.h, &flye.m);
int sun = 0, night = 0;
int tmp1 = cal(sunup.h, sunup.m);
int tmp2 = cal(sundown.h, sundown.m);
int tp1 = cal(flys.h, flys.m);
int tp2 = cal(flye.h, flye.m);
int time = tp2 - tp1;
if(time > 120)
flag = 1;
if(tp2 < tmp1)
night += (tp2 - tp2);
else if(tp2 > tmp1 && tp2 <= tmp2 && tp1 <= tmp1)
{
sun += (tp2 - tmp1);
night += (tmp1 - tp1);
}
else if(tp1 >= tmp1 && tp2 <= tmp2)
sun += (tp2 - tp1);
else if(tp1 <= tmp2 && tp1 >= tmp1 && tp2 >= tmp2)
{
sun += (tmp2 - tp1);
night += (tp2 - tmp2);
}
else if(tp1 >= tmp2)
sun += (tp2 - tp1);
if(sun <= night)
{
night = tp2 - tp1;
sun = 0;
}
SUMS += sun;
SUMS += night;
}
if(flag == 1)
printf("NON\n");
else if(SUMS >= limit1)printf("PASS\n");
else printf("NON\n");
}
return 0;
}
I. Class Packing
I题就是贪心，因为当两个年纪的人数不相同的时候，限制要求是为人数少的为限制人数。所以我们就先满足低年级
人数的班数，如果不够就从上一年级借。这样一直借下去。对于六年级的就不用借了直接算就可以了。今天写晕了，
最后重写才过的:
代码：
#include <iostream>
#include <cstdio>
#include <string>
#include <string.h>
#include <map>
#include <vector>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <queue>
#include <set>
#include <stack>
#include <functional>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <cassert>
#include <bitset>
#include <stack>
#include <ctime>
#include <list>
#define INF 0x7fffffff
#define max3(a,b,c) (max(a,b)>c?max(a,b):c)
#define min3(a,b,c) (min(a,b)<c?min(a,b):c)
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
int QuickMod(int a,int b,int n)
{
int r = 1;
while(b)
{
if(b&1)
r = (r*a)%n;
a = (a*a)%n;
b >>= 1;
}
return r;
}
struct node
{
int con;
int fp;
} num[7];
int main()
{
while(scanf("%d%d%d%d%d%d%d", &num[0].con, &num[1].con, &num[2].con, &num[3].con, &num[4].con, &num[5].con, &num[6].con))
{
if(!num[0].con && !num[1].con && !num[2].con && !num[3].con && !num[4].con && !num[5].con && !num[6].con)break;
num[0].fp = 20, num[1].fp = 20, num[2].fp = 20;
num[3].fp = 25, num[4].fp = 25;
num[5].fp = 30, num[6].fp = 30;
int ans = 0;
for(int i = 0; i < 7; ++i)
{
int tp1 = num[i].con, tp2 = num[i].fp;
if(i < 6)
{
ans += (tp1 / tp2);
if(tp1 % tp2)
{
ans++;
int tp = tp1 % tp2;
num[i+1].con -= (tp2 - tp);
if(num[i+1].con < 0)
num[i+1].con = 0;
}
}
else
{
ans += (tp1 / tp2);
if(tp1 % tp2)
ans++;
}
}
printf("%d\n", ans);
}
return 0;
}
J. Esspe-Peasee
扩展欧几里德，求解A*x+B*y == C,给出A,B,C求出X+Y的最小值。谔谔以为贪心，欧几里德，还有那个格式需要注
意，1的时候是不需要加s的。这里估计会卡卡.
学习一个代码：
#include <iostream>
#include <cstdio>
#include <string>
#include <string.h>
#include <map>
#include <vector>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <queue>
#include <set>
#include <stack>
#include <functional>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <cassert>
#include <bitset>
#include <stack>
#include <ctime>
#include <list>
#define INF 0x7fffffff
#define max3(a,b,c) (max(a,b)>c?max(a,b):c)
#define min3(a,b,c) (min(a,b)<c?min(a,b):c)
#define mem(a,b) memset(a,b,sizeof(a))
#define LL long long
using namespace std;
LL gcd(LL a, LL b)
{
return b == 0 ? a : gcd(b, a % b);
}
void kuozhan(LL a, LL b, LL &d, LL& x, LL& y)
{
if(!b)
{
d = a;
x = 1, y = 0;
}
else
{
kuozhan(b, a % b, d, y, x);
y -= x*(a / b);
}
}
int main()
{
LL x, y, a, b, c, d, t;
while(scanf("%lld%lld%lld", &a, &b, &c))
{
if(!a && !b && !c)
break;
LL g = gcd(a, b);
if(c % g)
cout << "Unquibable!" << endl;
else
{
a /= g, b /= g, c /= g;
kuozhan(a, b, d, x, y);
t = c % b;
x %= b, x = (x * t) % b, x = (x + b) % b, y = (c - (x * a)) / b;
if(y < 0)
{
printf("Unquibable!\n");
continue;
}
if(x == 1)
{
if(y == 1)printf("1 foom and 1 foob");
else printf("1 foom and %lld foobs",y);
}
else
{
if(y == 1)printf("%lld fooms and 1 foob",x);
else
printf("%lld fooms and ",x), printf("%lld foobs",y);
}
printf(" for a twob!\n");
}
}
return 0;
}