Codeforces Round #279 (Div. 2) Problem solving report_html/css_WEB-ITnose

A - Team Olympiad

Greedy question. . Just start from the first one.

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 __int64int a[6000], b[6000];int main(){    int n, x, y, z, ans, i, j;    while(scanf("%d",&n)!=EOF)    {        x=y=z=0;        for(i=0;i<n scanf memset for if x b else y z ans="min(x,min(y,z));" printf puts return> <br> B - Queue <br> <p></p> <p>By recording the next position in an array, the even numbers are positions and filling in odd positions. </p> <p>Even-numbered positions must start from 0, and odd-numbered positions must start from a number with no out-degree and only in-degree. </p> <p>The code is as follows: </p> <p></p> <pre name="code" class="sycode">#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 __int64int next[1100000], a[1100000], out[1100000];struct node{    int u, v;}fei[1100000];int main(){    int n, i, j, u, v, cnt, pos;    while(scanf("%d",&n)!=EOF)    {        memset(next,-1,sizeof(next));        memset(out,0,sizeof(out));        for(i=0;i<n scanf next out cnt="1;" for if break a pos="fei[i].u;" printf return> <br> C - Hacking Cypher <br> <p></p> <p>Scan the record from the front and back respectively to be divisible location. Very good handling from front to back. </p> <p>As for the k-th digit, you can first preprocess the sum of the remainders of 10^k to b and the remainder of the first k digits to b, and then, the last few digits = total number - the previous The number represented by k bits*10^k. So as long as the total number %b==(the number represented by the first k digits %b)*(10^k%b)%b is satisfied, it means that the number represented by the last few digits can be divided by b. </p> <p>This can be completed within O(n) complexity. </p> <p>It means that I am really weak. . I’ve seen other people make it in a short time. . But it took me half an hour to figure it out. . . </p> <p>The code is as follows: </p> <p></p> <pre name="code" class="sycode">#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 __int64int a[1100000], b[1100000], c[1100000], d[1100000];char s[1100000];int main(){    int aa, bb, i, x, len, pos, flag=0, y;    gets(s);    scanf("%d%d",&aa,&bb);    memset(a,0,sizeof(a));    memset(d,0,sizeof(d));    x=1;    c[1]=1%bb;    for(i=2;i <br> D - Chocolate <br> <p></p> <p>If the final areas are equal, then the factor of 2 The number must be equal to the number of factors of 3. So we can first find the number of factors of 2 and the number of factors of 3. Then at this time we can have two operations: eliminating a 2 or turning a 3 into a 2. So at this time, first change the larger of 3 into 2 to make the remaining 3 equal, and then eliminate the larger factor of 2. The larger side makes equal. Then it is judged that the areas of both sides are equal at this time. </p> <p>The code is as follows: </p> <p></p> <pre name="code" class="sycode">#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 __int64int main(){    int a1, b1, a2, b2, x2, x3, y2, y3, ans, m1, m2, n1, n2;    while(scanf("%d%d%d%d",&a1,&b1,&a2,&b2)!=EOF)    {        m1=a1;        m2=a2;        n1=b1;        n2=b2;        x2=x3=y2=y3=0;        while(!(a1%2)||!(a1%3))        {            if(a1%2==0)            {                x2++;                a1/=2;            }            if(a1%3==0)            {                x3++;                a1/=3;            }        }        while(!(b1%2)||!(b1%3))        {            if(b1%2==0)            {                x2++;                b1/=2;            }            if(b1%3==0)            {                x3++;                b1/=3;            }        }        while(!(a2%2)||!(a2%3))        {            if(a2%2==0)            {                y2++;                a2/=2;            }            if(a2%3==0)            {                y3++;                a2/=3;            }        }        while(!(b2%2)||!(b2%3))        {            if(b2%2==0)            {                y2++;                b2/=2;            }            if(b2%3==0)            {                y3++;                b2/=3;            }        }        if(x3>=y3)        {            x2+=x3-y3;            ans=x3-y3;            for(int i=0; i<x3-y3 i if m1="m1/3*2;" else n1="n1/3*2;" y2 ans="y3-x3;" for m2="m2/3*2;" n2="n2/3*2;">=y2)        {            for(int i=0; i<x2-y2 i if m1 else n1 for m2 n2 puts printf return> <br> <br> <p></p> </x2-y2></x3-y3></algorithm></set></map></queue></ctype.h></math.h></stdlib.h></cstring></string></cstdio></iostream>