2017年浙江工业大学大学生程序设计迎新赛预赛

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A - 栗酱的亦或和

nim游戏

#include <iostream>#include <string.h>#include <stdio.h>#include <algorithm>using namespace std;const int N = 1e5 + 10;int T,x,y,z,n,m;int a[N];int main(){    scanf("%d",&T);    while(T--){        scanf("%d%d",&n,&m);        int sum = 0;        for(int i=0;i<n;i++){            scanf("%d",&a[i]);            if(i==m-1) continue;            if(i) sum ^= a[i];            else sum = a[i];        }        if(sum==0) printf("Yes\n");        else{            if(sum<a[m-1]) printf("Yes\n");            else printf("No\n");        }    }    return 0;}

D - 简单的数据结构

模拟

#include <iostream>#include <string.h>#include <stdio.h>#include <algorithm>using namespace std;const int N = 1e6 + 10;bool cmp(int a,int b){    return a>b;}int n,m,a,ope,st[N],be=50000+10,ed=50000+9,re=0,num=0;int main(){    scanf("%d%d",&n,&m);    for(int i=0;i<m;i++){        scanf("%d",&ope);        if(ope==1||ope==3){            scanf("%d",&a);            num++;            if(ope==1){                if(re) st[++ed] = a;                else st[--be] = a;            }else{                if(!re) st[++ed] = a;                else st[--be] = a;            }        }else{            if(!re){                switch(ope){                    case 2:be++; num--; break;                    case 4:ed--; num--; break;                    case 5:re = !re; break;                    case 6:                        printf("%d\n",num);                        for(int i=be;i<=ed;i++)                            printf("%d%c",st[i],i==ed?'\n':' ');                        break;                    case 7:sort(st+be,st+ed+1); break;                }            }else{                switch(ope){                    case 4:be++; num--; break;                    case 2:ed--; num--; break;                    case 5:re = !re; break;                    case 6:                        printf("%d\n",num);                        for(int i=ed;i>=be;i--)                            printf("%d%c",st[i],i==be?'\n':' ');                        break;                    case 7:sort(st+be,st+ed+1,cmp); break;                }            }        }    }    return 0;}

I - 栗酱数数

暴力

#include <iostream>#include <string.h>#include <stdio.h>using namespace std;int main(){    int n;        scanf("%d",&n);        for(int i=1; i<=n; i++){        if(!(i%4)) continue;        int now = i, flag = 0;        while(now){            if(now%10==4) {                flag = 1;                break;            }            now /= 10;        }        if(flag) continue;                printf("%d\n",i);    }    return 0;}

J - 裁缝大师

极坐标

#include <iostream>#include <string.h>#include <stdio.h>#include <algorithm>#include <math.h>using namespace std;const int N = 1e5 + 10;const double PI = 4 * atan(1.0);int T;double x,y,z,n,m;int a[N];int main(){    scanf("%d",&T);    while(T--){        scanf("%lf%lf%lf%lf",&x,&y,&z,&n);        double ang1 = 2 * PI / n;        double p = 0;        double x0,y0;        x0 = x+z; y0 = y;        printf("%.2lf %.2lf\n",x0,y0);        for(int i=1;i<n;i++){            p -= ang1;            x0 = z*cos(p)+x; y0 = z*sin(p)+y;            if(fabs(x0)<0.01) x0 = 0;            if(fabs(y0)<0.01) y0 = 0;            printf("%.2lf %.2lf\n",x0,y0);        }            }    return 0;}

K - 栗酱的连通图

贪心

#include <iostream>#include <string.h>#include <stdio.h>#include <algorithm>using namespace std;const int N = 1e3 + 10;int T,x,y,z,n,m;int a[N];int main(){    scanf("%d",&T);    while(T--){        int sum = 0, mx = 0;        scanf("%d",&n);        for(int i=0;i<n;i++){            scanf("%d",&a[i]);            sum += a[i]/2;            mx = max(mx, a[i]/2);        }        sum += (n-2) * mx;        printf("%d\n",sum);    }    return 0;}

L - 取数游戏

#include <iostream>#include <string.h>#include <stdio.h>#include <algorithm>using namespace std;const int N = 1e2 + 10;int T,x,y,z,n,m;int a[N];int main(){    scanf("%d",&T);    while(T--){        scanf("%d%d%d",&a[0],&a[1],&a[2]);        sort(a,a+3);        printf("%d%d%d\n",a[2],a[1],a[0]);    }    return 0;}

M - 小杰的签到题

贪心

#include <iostream>#include <string.h>#include <stdio.h>#include <algorithm>using namespace std;const int N = 1e3 + 10;int T,x,y,z,n,m;int a[N];int main(){    scanf("%d",&T);    while(T--){        scanf("%d",&n);        for(int i=0;i<n;i++){            scanf("%d",&a[i]);        }        scanf("%d",&m);        sort(a,a+n);        x = y = z = -1;        for(int i=0;i<n;i++){            if(x<=a[i]) x = a[i] + m;            else if(y<=a[i]) y = a[i] + m;            else if(z<=a[i]) z = a[i] + m;            else {                if(x<=y&&x<=z) x += m;                else if(y<=x&&y<=z) y += m;                else z += m;            }        }        printf("%d\n",max(x,max(y,z)));    }    return 0;}

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