Codeforces Round #280 (Div. 2) Solution Report A.B.C.D.E._html/css_WEB-ITnose

I don’t know whether it’s because my level has improved or because the CF questions have changed. . . . . .

A - Vanya and Cubes

Water question. . Violent enumeration will do. .

The code is as follows:

#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <stdlib.h>#include <math.h>#include <ctype.h>#include <queue>#include <map>#include <set>#include <algorithm>using namespace std;#define LL __int64const int INF=0x3f3f3f3f;int s[100], sum[100];int main(){        int n, i;        s[1]=1;        sum[i]=1;        scanf("%d",&n);        for(i=2;i<=10000;i++){                s[i]=s[i-1]+i;                sum[i]=sum[i-1]+s[i];                if(sum[i]>=n)break;        }        printf("%d\n",i-1);        return 0;}

B - Vanya and Lanterns

Water question. . Violently find the shortest distance/2 between two adjacent ones, and then compare it with the boundary to find the longest distance.

The code is as follows:

#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <stdlib.h>#include <math.h>#include <ctype.h>#include <queue>#include <map>#include <set>#include <algorithm>using namespace std;#define LL __int64const int INF=0x3f3f3f3f;int a[2000];int main(){        int n, l, i, j;        double d, max1=-1;        scanf("%d%d",&n,&l);        a[0]=0;        a[n+1]=l;        for(i=1;i<=n;i++){                scanf("%d",&a[i]);        }        sort(a,a+n+2);        for(i=2;i<=n;i++){                max1=max(max1,(a[i]-a[i-1])/2.0);        }        max1=max(max1,a[1]-0.0);        max1=max(max1,l*1.0-a[n]);        printf("%.10lf\n",max1);        return 0;}

C - Vanya and Exams

Greedy.

Start with the fewest papers required and add up.

The code is as follows:

#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <stdlib.h>#include <math.h>#include <ctype.h>#include <queue>#include <map>#include <set>#include <algorithm>using namespace std;#define LL __int64const int INF=0x3f3f3f3f;struct node{        LL f, e;}fei[110000];int cmp(node x, node y){        return x.e<y.e;}int main(){        int i;        LL n, r, ave, s, sum, ss, ans=0;        scanf("%I64d%I64d%I64d",&n,&r,&ave);        sum=n*ave;        s=0;        for(i=0;i<n;i++){                scanf("%I64d%I64d",&fei[i].f,&fei[i].e);                s+=fei[i].f;        }        ss=sum-s;        if(ss<=0)        {                printf("0\n");                return 0;        }        sort(fei,fei+n,cmp);        //printf("%d\n",ss);        for(i=0;i<n;i++){                if(fei[i].f>=r) continue ;                if(ss-(r-fei[i].f)<=0){                        ans+=ss*fei[i].e;                        break;                }                else{                        ss-=r-fei[i].f;                        ans+=fei[i].e*(r-fei[i].f);                }                //printf("%d\n",ss);        }        printf("%I64d\n",ans);        return 0;}

D - Vanya and Computer Game

I thought of the two-point time during the game. I have never figured out how to tell who it is after finding the time. . It dawned on me the next day. . Just judge whether it is divisible or not. . . . Why didn't I think of that at the time? . . depressed. .

First zoom in on the time axis, zoom in 1/x and 1/y seconds to y and x, and then divide the time into two halves. After finding the time, judge whether it is divisible.

The code is as follows:

#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <stdlib.h>#include <math.h>#include <ctype.h>#include <queue>#include <map>#include <set>#include <algorithm>using namespace std;#define LL __int64const int INF=0x3f3f3f3f;int main(){        LL n, x, y, i, z, low, high, mid, ans;        scanf("%I64d%I64d%I64d",&n, &x, &y);        while(n--) {                scanf("%I64d",&z);                low=1;                high=1e16;                while(low<=high) {                        mid=low+high>>1;                        if(mid/x+mid/y>=z) {                                high=mid-1;                                ans=mid;                        }                        else low=mid+1;                }                if(ans%x==0&&ans%y==0) puts("Both");                else if(ans%x==0) puts("Vova");                else puts("Vanya");        }        return 0;}

E - Vanya and Field

I feel like the E question this time is good. . . First of all, because gcd is 1, it can be guaranteed that all n numbers can be accessed and will not be repeated. So you can let him start from a certain point with the abscissa 0 by default. Then when the ordinate is 0, a mapping relationship between the x coordinate and the y coordinate can be found and saved. Then every time you enter a coordinate, you can use a formula to determine which point it starts from. Then just find the most starting point.

The code is as follows:

#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <stdlib.h>#include <math.h>#include <ctype.h>#include <queue>#include <map>#include <set>#include <algorithm>using namespace std;#define LL __int64const int INF=0x3f3f3f3f;int _hash[1100000], a[1100000];int main(){        int n, m, dx, dy, i, j, x, y, pos, max1;        scanf("%d%d%d%d",&n,&m,&dx,&dy);        x=y=0;        for(i=0;i<n;i++){                a[x]=y;                x=(x+dx)%n;                y=(y+dy)%n;        }        memset(_hash,0,sizeof(_hash));        while(m--){                scanf("%d%d",&x,&y);                _hash[(y+n-a[x])%n]++;        }        max1=-1;        for(i=0;i<n;i++){                if(max1<_hash[i]){                        max1=_hash[i];                        pos=i;                }        }        printf("0 %d\n",pos);        return 0;}