Codeforces Round #452 (Div. 2) A B C
Codeforces Round #452 (Div. 2)
A Splitting in Teams
题目链接：
http://codeforces.com/contest/899/problem/A
思路：
统计1和2出现的次数，尽量使2能够与1匹配尽可能多用。假设1再匹配完2之后还有剩余，则求出3个1可组成的方案
代码：
#include
using namespace std;
typedef long long ll;
const int maxn = 200005;
ll num,cnt[3];
int main() {
ll n;
scanf("%I64d",&n);
for(ll i=0;i B Proper Nutrition
题目链接：
http://codeforces.com/contest/899/problem/B
思路：
暴力枚举4种情况，24可以跨越三年，谨记...
代码：
#include
using namespace std;
int b[4][36]={
{31,28,31,30,31,30,31,31,30,31,30,31,31,28,31,30,31,30,31,31,30,31,30,31,31,28,31,30,31,30,31,31,30,31,30,31},
{31,28,31,30,31,30,31,31,30,31,30,31,31,28,31,30,31,30,31,31,30,31,30,31,31,29,31,30,31,30,31,31,30,31,30,31},
{31,28,31,30,31,30,31,31,30,31,30,31,31,29,31,30,31,30,31,31,30,31,30,31,31,28,31,30,31,30,31,31,30,31,30,31},
{31,29,31,30,31,30,31,31,30,31,30,31,31,28,31,30,31,30,31,31,30,31,30,31,31,28,31,30,31,30,31,31,30,31,30,31}
};
int a[24];
int main() {
int n;
scanf("%d",&n);
for(int i=0;i C Phone Numbers
题目链接：
http://codeforces.com/contest/899/problem/C
思路：
我们尽可能让两组都尽可能趋近于sum/2。因为给出的n个数字是从1到n连续的，所以利用这个性质，假定其中较小的一组必定是连续的，标记为1。输出没有被标记的即可。特判一下n=2的情况。
代码：
#include
using namespace std;
typedef long long ll;
int a[60005];
int main() {
ll n;
ll dis=0;
ll cnt=0;
scanf("%I64d",&n);
if(n==2) {
printf("1\n");
printf("1 1\n");
return 0;
}
memset(a,0,sizeof(a));
ll sum=(1+n)*n/2;
sum/=2;
for(ll i=1;i<=n;++i) {
if(sum%(i+i+1)==0) {
ll temp=sum/(i+i+1);
if(temp<=i) {
cnt=n-temp*2;
for(ll j=i-temp+1;j<=i+temp;++j) {
a[j]=1;
dis+=j;
}
break;
}
}
}
printf("%I64d\n",abs((1+n)*n/2-dis-dis));
printf("%I64d ",cnt);
int index=0;
for(int i=1;i<=n;++i) {
if(a[i]==0) {
if(index==0) printf("%d",i);
else printf(" %d",i);
++index;
}
}
printf("\n");
return 0;
}