2015 Multi-University Training Contest 5
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官方题解:2015 Multi-University Training Contest 5 solutions BY 绍兴一中
1002
题意:给出一个长度为\(n\)递推数列\(A\),求出所有\((A_i+A_j),(1\leqslant i,j\leqslant n)\)相异或的结果。
注意\(i,j\)之间没有关系限制,也就是说可以\(i==j\)
因为\((A_i+A_j) xor (A_j+A_i)==0\)
所以就只用计算所有\((A_i+A_i) \)相异或的值
#include<cstdio>#include<algorithm>#include<iostream>#include<cmath>#include<cstring>#include<queue>#include<vector>#include<string>using namespace std;#define MAX(a,b) ((a>b)?(a):(b))#define MIN(a,b) ((a<b)?(a):(b))#define LL __int64#define N 500005#define INF 1<<30LL a[N];int main(){int T;scanf("%d",&T);LL n,m,z,l;while(T--){ cin>>n>>m>>z>>l; a[1]=0; LL ans=0; for(int i=2;i<=n;++i){ a[i]=(a[i-1]*m+z)%l; ans^=(a[i]*2); } cout<<ans<<endl;}return 0;}
1005
题意:定义\(F(x)\)为原子系数为x的化学元素的第一电离能,给出\(u,v\),比较\(F(u),F(v)\)的大小。
直接打表。
#include<cstdio>#include<algorithm>#include<iostream>#include<cmath>#include<cstring>#include<queue>#include<vector>#include<string>using namespace std;#define MAX(a,b) ((a>b)?(a):(b))#define MIN(a,b) ((a<b)?(a):(b))#define LL __int64#define N 105#define INF 1<<30double f[N]={0,1312.0,2372.3,520.2,899.5,800.6,1086.5,1402.3,1313.9,1681.0,2080.7,495.8,737.7,577.5,786.5,1011.8,999.6,1251.2,1520.6};int main(){ f[35]=1139.9; f[36]=1350.8; f[53]=1008.4; f[54]=1170.4; f[85]=850.0; f[86]=1037.0;int u,v;while(~scanf("%d%d",&u,&v)){ printf("%s\n",(f[u]>f[v]?"FIRST BIGGER":"SECOND BIGGER"));}return 0;}
1007
HDU 5349:http://acm.hdu.edu.cn/showproblem.php?pid=5349
题意:给出一个multiset和三种操作:
1 x : 将数 \(x\) 加入集合中
2 : 删除一个集合中最小的数 (如果集合为空,忽略此操作)
3 : 询问集合中最大的数是多少 (如果集合为空,最大值为0)
直接用multiset可水过
#include<cstdio>#include<algorithm>#include<iostream>#include<cmath>#include<cstring>#include<queue>#include<vector>#include<map>#include<set>using namespace std;#define MAX(a,b) a>b?a:b#define MIN(a,b) a<b?a:b#define N 100005#define INF 1<<30int main(){ int n; multiset<int> s; multiset<int> ::iterator it; int op,x; while(scanf("%d",&n)!=EOF){ while(n--){ scanf("%d",&op); if(op==1){ scanf("%d",&x); s.insert(x); } else if(op==2){ if(!s.empty()) s.erase(s.begin()); } else{ if(!s.empty()){ it=--s.end(); printf("%d\n",(*it)); } else{ puts("0"); } } } } return 0;}
(待续。。。)
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