2016 Multi-University Training Contest 2
2016 Multi-University Training Contest 2
5734 Acperience
直接队友代码
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <sstream> 5 #include <string> 6 #include <algorithm> 7 #include <list> 8 #include <map> 9 #include <vector> 10 #include <queue> 11 #include <stack> 12 #include <cmath> 13 #include <cstdlib> 14 using namespace std; 15 long long a[100006]; 16 int main() 17 { 18 //freopen("in.txt","r",stdin); 19 int T; 20 scanf("%d",&T); 21 while(T--) 22 { 23 long long p = 0,q; 24 memset(a,0,sizeof(a)); 25 int n; 26 long long sum = 0; 27 long long sum2 = 0; 28 scanf("%d",&n); 29 for(int i = 0; i < n; i ++) 30 { 31 scanf("%I64d",&a[i]); 32 if(a[i] < 0) 33 a[i] = -a[i]; 34 sum += a[i]; 35 sum2 += a[i] * a[i]; 36 } 37 sum = sum * sum; 38 p = sum2* n- sum; 39 40 q = n; 41 int temp = __gcd(p,q); 42 p = p / temp; 43 q = q/ temp; 44 printf("%I64d/%I64d\n",p,q); 45 46 } 47 return 0; 48 }
5724 It's All In The Mind
水
1 // #pragma comment(linker, "/STACK:102c000000,102c000000") 2 #include <iostream> 3 #include <cstdio> 4 #include <cstring> 5 #include <sstream> 6 #include <string> 7 #include <algorithm> 8 #include <list> 9 #include <map> 10 #include <vector> 11 #include <queue> 12 #include <stack> 13 #include <cmath> 14 #include <cstdlib> 15 // #include <conio.h> 16 using namespace std; 17 #define clc(a,b) memset(a,b,sizeof(a)) 18 #define inf 0x3f3f3f3f 19 #define lson l,mid,rt<<1 20 #define rson mid+1,r,rt<<1|1 21 const int N = 1e5+10; 22 const int M = 1e6+10; 23 const int MOD = 1e9+7; 24 #define LL long long 25 #define mi() (l+r)>>1 26 double const pi = acos(-1); 27 28 void fre() { 29 freopen("in.txt","r",stdin); 30 } 31 32 // inline int r() { 33 // int x=0,f=1;char ch=getchar(); 34 // while(ch>'9'||ch<'0') {if(ch=='-') f=-1;ch=getchar();} 35 // while(ch>='0'&&ch<='9') { x=x*10+ch-'0';ch=getchar();}return x*f; 36 // } 37 38 int a[110]; 39 int main(){ 40 int T; 41 scanf("%d",&T); 42 while(T--){ 43 clc(a,0); 44 int n,m; 45 scanf("%d%d",&n,&m); 46 int sum=0; 47 int last=2; 48 for(int i=0;i<m;i++){ 49 int pos; 50 scanf("%d",&pos); 51 scanf("%d",&a[pos]); 52 if(pos==1||pos==2){ 53 sum+=a[pos]; 54 continue; 55 } 56 int len=pos-last; 57 sum+=len*a[pos]; 58 last=pos; 59 } 60 // cout<<sum<<endl; 61 if(!a[1]) a[1]=100,sum+=100; 62 if(!a[2]) a[2]=a[1],sum+=a[2]; 63 // cout<<sum<<endl; 64 int g=__gcd(sum,a[1]+a[2]); 65 printf("%d/%d\n",(a[1]+a[2])/g,sum/g); 66 } 67 return 0; 68 }
5745 La Vie en rose
题意有好难理解，其实就是只能交换相邻字符且一次
队友代码
1 #include <cstdio> 2 #include <cstring> 3 #include <iostream> 4 #include <algorithm> 5 #include <vector> 6 #include <queue> 7 #include <set> 8 #include <map> 9 #include <string> 10 #include <cmath> 11 #include <stdlib.h> 12 using namespace std; 13 typedef long long LL; 14 const int inf=0x3f3f3f3f; 15 const int N=1e5+10; 16 char s1[N],s2[5010]; 17 bool ans[N]; 18 int main() 19 { 20 //freopen("in.txt","r",stdin); 21 int t,n,m; 22 scanf("%d",&t); 23 while(t--) 24 { 25 scanf("%d%d",&n,&m); 26 scanf("%s%s",s1,s2); 27 memset(ans,0,sizeof(ans)); 28 for(int i=0;i<n;i++) 29 { 30 int p=i,j; 31 for(j=0;j<m;j++) 32 { 33 if(s1[p+j]==s2[j]) continue; 34 if(s1[p+j]==s2[j+1]&&s1[p+j+1]==s2[j]) j++; 35 else break; 36 } 37 if(j>=m) ans[i]=1; 38 } 39 for(int i=0;i<n;i++) 40 printf("%d",ans[i]); 41 printf("\n"); 42 } 43 return 0; 44 }
5744 Keep On Movin
把一堆字母分成多组回文串，求每组里面最短的最长
水
1 // #pragma comment(linker, "/STACK:102c000000,102c000000") 2 #include <iostream> 3 #include <cstdio> 4 #include <cstring> 5 #include <sstream> 6 #include <string> 7 #include <algorithm> 8 #include <list> 9 #include <map> 10 #include <vector> 11 #include <queue> 12 #include <stack> 13 #include <cmath> 14 #include <cstdlib> 15 // #include <conio.h> 16 using namespace std; 17 #define clc(a,b) memset(a,b,sizeof(a)) 18 #define inf 0x3f3f3f3f 19 #define lson l,mid,rt<<1 20 #define rson mid+1,r,rt<<1|1 21 const int N = 1e5+10; 22 const int M = 1e6+10; 23 const int MOD = 1e9+7; 24 #define LL long long 25 #define mi() (l+r)>>1 26 double const pi = acos(-1); 27 28 void fre() { 29 freopen("in.txt","r",stdin); 30 } 31 32 // inline int r() { 33 // int x=0,f=1;char ch=getchar(); 34 // while(ch>'9'||ch<'0') {if(ch=='-') f=-1;ch=getchar();} 35 // while(ch>='0'&&ch<='9') { x=x*10+ch-'0';ch=getchar();}return x*f; 36 // } 37 38 int main(){ 39 // fre(); 40 int T; 41 int n; 42 scanf("%d",&T); 43 while(T--){ 44 scanf("%d",&n); 45 int num=0,cnt=0; 46 LL sum=0; 47 for(int i=0;i<n;i++){ 48 int x; 49 scanf("%d",&x); 50 sum+=x; 51 if(x&1){ 52 num++; 53 cnt+=x/2; 54 } 55 else 56 cnt+=x/2; 57 } 58 if(num==0){ 59 printf("%I64d\n",sum); 60 continue; 61 } 62 int ans; 63 ans=1+cnt/num*2; 64 printf("%d\n",ans); 65 } 66 return 0; 67 }
5738 Eureka
题意：求多少个子集共线
思路：先按x，y排序个。从左边开始枚举每个点当作起点，统计右边有多少个共线的子集。用极坐标排序，但是判断点共线的时候把斜率转换成相乘的形式。
有n个点和i共线的话，那么子集为2^n-1,注意重点的情况会多算d-1次
1 // #pragma comment(linker, "/STACK:102c000000,102c000000") 2 #include <iostream> 3 #include <cstdio> 4 #include <cstring> 5 #include <sstream> 6 #include <string> 7 #include <algorithm> 8 #include <list> 9 #include <map> 10 #include <vector> 11 #include <queue> 12 #include <stack> 13 #include <cmath> 14 #include <cstdlib> 15 // #include <conio.h> 16 using namespace std; 17 #define clc(a,b) memset(a,b,sizeof(a)) 18 #define inf 0x3f3f3f3f 19 #define lson l,mid,rt<<1 20 #define rson mid+1,r,rt<<1|1 21 const int N = 1e5+10; 22 const int M = 1e6+10; 23 const int MOD = 1e9+7; 24 #define LL long long 25 #define mi() (l+r)>>1 26 double const pi = acos(-1); 27 const double eps = 1e-8; 28 void fre() { 29 freopen("in.txt","r",stdin); 30 } 31 32 // inline int r() { 33 // int x=0,f=1;char ch=getchar(); 34 // while(ch>'9'||ch<'0') {if(ch=='-') f=-1;ch=getchar();} 35 // while(ch>='0'&&ch<='9') { x=x*10+ch-'0';ch=getchar();}return x*f; 36 // } 37 38 struct Point{ 39 int x,y; 40 double ang; 41 }p[1010],p2[1010]; 42 43 bool cmp(Point a,Point b){ 44 if(a.y==b.y) return a.x<b.x; 45 return a.y<b.y; 46 } 47 48 bool cmp2(Point a,Point b){ 49 return a.ang<b.ang; 50 } 51 int f[1010]; 52 int main(){ 53 // fre(); 54 f[0]=1; 55 for(int i=1;i<1005;i++){ 56 f[i]=(f[i-1]*2)%MOD; 57 } 58 int T; 59 scanf("%d",&T); 60 while(T--){ 61 int n; 62 scanf("%d",&n); 63 for(int i=1;i<=n;i++){ 64 scanf("%d%d",&p[i].x,&p[i].y); 65 } 66 LL ans=0; 67 sort(p+1,p+1+n,cmp); 68 for(int i=1;i<n;i++){ 69 int s=0,cnt=0; 70 for(int j=i+1;j<=n;j++){ 71 if(p[j].x==p[i].x&&p[j].y==p[i].y) {s++;continue;} 72 p2[++cnt].ang=atan2(p[j].y-p[i].y,p[j].x-p[i].x); 73 p2[cnt].x=p[j].x; 74 p2[cnt].y=p[j].y; 75 } 76 sort(p2+1,p2+1+cnt,cmp2); 77 int k,d=0; 78 for(int j=1;j<=cnt;){ 79 int num=s+1; 80 for(k=j+1;k<=cnt;k++){ 81 if((p2[k].y-p2[j].y)*(p2[j].x-p[i].x)!=(p2[j].y-p[i].y)*(p2[k].x-p2[j].x)) break; 82 num++; 83 } 84 j=k; 85 d++; 86 ans=(ans+f[num]-1)%MOD; 87 } 88 ans=(ans-(LL)(d-1)*(f[s]-1)%MOD+MOD)%MOD; 89 } 90 printf("%I64d\n",ans); 91 } 92 return 0; 93 }
