CodeForces 560A，B，C

CodeForces 560A

题意：给定一个货币系统，问不能组成的最小的钱数是多少。

思路：水，只要检查有没有出现1即可，有则输出-1，否则1.

code：

#include <iostream>#include <cstdio>#include <cmath>#include <algorithm>#include <cstring>#include <sstream>#include <string>#include <vector>#include <list>#include <queue>#include <stack>#include <map>#include <set>#include <bitset>using namespace std;typedef long long ll;typedef unsigned long long ull;typedef long double ld;const int INF=0x3fffffff;const int inf=-INF;const int N=1000000+5;const int M=1e3+5;const int mod=1000000007;const double pi=acos(-1.0);#define cls(x,c) memset(x,c,sizeof(x))#define cpy(x,a) memcpy(x,a,sizeof(a))#define ft(i,s,n) for (int i=s;i<=n;i++)#define frt(i,s,n) for (int i=s;i>=n;i--)#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define lrt  rt<<1#define rrt  rt<<1|1#define middle int m=(r+l)>>1#define lowbit(x) (x&-x)#define pii pair<int,int>#define mk make_pair#define IN freopen("in.txt","r",stdin);#define OUT freopen("out.txt","w",stdout);int v[M],vis[N];int main(){    int n;cls(vis,0);    scanf("%d",&n);    ft(i,1,n) scanf("%d",&v[i]),vis[v[i]]=1;    sort(v+1,v+1+n);    if (v[1]==1) puts("-1");    else puts("1");}

CodeForces 560B

题意：给定一个相框，问能否把两张照片放进去。。

思路：依旧水，对于两个照片，只有长长，短短，长短和短长这四种放法，没swap会wa？？

code：

#include <iostream>#include <cstdio>#include <cmath>#include <algorithm>#include <cstring>#include <sstream>#include <string>#include <vector>#include <list>#include <queue>#include <stack>#include <map>#include <set>#include <bitset>using namespace std;typedef long long ll;typedef unsigned long long ull;typedef long double ld;const int INF=0x3fffffff;const int inf=-INF;const int N=1000000;const int M=2005;const int mod=1000000007;const double pi=acos(-1.0);#define cls(x,c) memset(x,c,sizeof(x))#define cpy(x,a) memcpy(x,a,sizeof(a))#define ft(i,s,n) for (int i=s;i<=n;i++)#define frt(i,s,n) for (int i=s;i>=n;i--)#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define lrt  rt<<1#define rrt  rt<<1|1#define middle int m=(r+l)>>1#define lowbit(x) (x&-x)#define pii pair<int,int>#define mk make_pair#define IN freopen("in.txt","r",stdin);#define OUT freopen("out.txt","w",stdout);int a[5],b[5];bool ok(){    int s1,s2;    s1=max(b[3],b[2]);    s2=a[3]+a[2];    if (s1>s2) swap(s1,s2);    if (s1<=a[1]&&s2<=b[1]) return 1;    s1=max(a[3],a[2]);    s2=b[3]+b[2];    if (s1>s2) swap(s1,s2);    if (s1<=a[1]&&s2<=b[1]) return 1;    s1=max(b[3],a[2]);    s2=a[3]+b[2];    if (s1>s2) swap(s1,s2);    if (s1<=a[1]&&s2<=b[1]) return 1;    s1=max(a[3],b[2]);    s2=b[3]+a[2];    if (s1>s2) swap(s1,s2);    if (s1<=a[1]&&s2<=b[1]) return 1;    return 0;}int main(){    ft(i,1,3) {        scanf("%d %d",&a[i],&b[i]);        if (a[i]>b[i]) swap(a[i],b[i]);    }    if(ok()) puts("YES");    else puts("NO");}

CodeForces 560C

题意：给定一个六边形的各边长度，已知相邻两边夹角均为120度，求能分成多少变长为1的正三角形。

思路：分别延长v2，v4，v6得到一个大的正三角形，用两个边的平方表示正三角形面积（多乘了2/根号3），然后减去多补的那部分。

code：

#include <iostream>#include <cstdio>#include <cmath>#include <algorithm>#include <cstring>#include <sstream>#include <string>#include <vector>#include <list>#include <queue>#include <stack>#include <map>#include <set>#include <bitset>using namespace std;typedef long long ll;typedef unsigned long long ull;typedef long double ld;const int INF=0x3fffffff;const int inf=-INF;const int N=1000000+5;const int M=1e5+5;const int mod=1000000007;const double pi=acos(-1.0);#define cls(x,c) memset(x,c,sizeof(x))#define cpy(x,a) memcpy(x,a,sizeof(a))#define ft(i,s,n) for (int i=s;i<=n;i++)#define frt(i,s,n) for (int i=s;i>=n;i--)#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define lrt  rt<<1#define rrt  rt<<1|1#define middle int m=(r+l)>>1#define lowbit(x) (x&-x)#define pii pair<int,int>#define mk make_pair#define IN freopen("in.txt","r",stdin);#define OUT freopen("out.txt","w",stdout);int v[8];int main(){    ft(i,1,6) scanf("%d",&v[i]);    int s=(v[1]+v[2]+v[3])*(v[1]+v[2]+v[3])-v[1]*v[1]-v[3]*v[3]-v[5]*v[5];    printf("%d\n",s);}