CF 301 Div2

A-combination Lock

The main topic: To have N (0-9) ring password lock, a few strings s1->s2 the minimum number of moves;

Topic Analysis:

Simple simulation;

Code:

const int N=100007;char s1[n],s2[n];int Main () {    int N;    while (scanf ("%d", &n) ==1)    {        int ans=0;        scanf ("%s%s", s1+1,s2+1);        for (int i=1;i<=n;i++)        {            int a=s1[i]-' 0 ';            int b=s2[i]-' 0 ';            if (a>b) swap (A, b);            Ans+=min (b-a,a+ (9-b) +1);        }        printf ("%d\n", ans);    }    return 0;}

B-school Marks

The main topic: N number, known as K, the median >=y; and all the number sum<=x; fill the remaining n-k number;

Topic Analysis:

Because sum is less than X, try to fill in decimals. Mid left Fill 1, right fill y;

Code:

vector<int> Ans;int Main () {int n,k,p,x,y;        while (scanf ("%d%d%d%d%d", &n,&k,&p,&x,&y) ==5) {ans.clear ();        int sum=0;        int mi=0;        int flag=0;            for (int i=0;i<k;i++) {int v;cin>>v;            if (v>=y) flag++;            if (v<y) mi++;        Sum+=v; } if (mi>n/2| |            sum>x) {printf (" -1\n");        Continue            } while (n/2+1-(K-MI) <0&&flag>1) {mi++;        flag--;            } for (int i=0;i<n/2-mi;i++) {ans.push_back (1);        Sum+=1;            } for (int i=0;i<n/2+1-(K-MI); i++) {ans.push_back (y);            Sum+=y;        flag=1; } if (sum<=x&&flag&& (signed int) ans.size () +k==n) {for (int i=0;i< (signed int) Ans.siz                E (); i++) {if (!i) printf ("%d", ans[i]);   else printf ("%d", ans[i]);         } printf ("\ n");        } else{printf (" -1\n"); }} return 0;}

C-ice Cave

The main idea: to the maze, X can not go, '. ' Walk two times, from (R1,C2)---> (r2,c2) request (R2,C2) walk two times;

Topic Analysis:

DFS,CNT records the number of times, (well, I was using the also mark, ' * ' means have gone over);

Code:

const int N=100007;char gra[555][555];int go[4][2]={{0,1},{0,-1},{-1,0},{1,0}};int r1,c1,r2,c2;int ok;int n,m;void Dfs (int cx,int cy)    {int x, y;    if (OK) return;        for (int i=0;i<4;i++) {x=cx+go[i][0];        Y=CY+GO[I][1]; if (1<=x&&x<=n&&1<=y&&y<=m) {if (gra[x][y]!= ' x ') {if (g                ra[x][y]== ' * ') {if (X==R2&AMP;&AMP;Y==C2) ok=1;                    } else{gra[x][y]= ' * ';                    DFS (x, y);                Gra[x][y]= '. ';        }}}}}int main () {while (scanf ("%d%d", &n,&m) ==2) {ok=0;        for (int i=1;i<=n;i++) scanf ("%s", &gra[i][1]);        scanf ("%d%d%d%d", &AMP;R1,&AMP;C1,&AMP;R2,&AMP;C2);        if (gra[r2][c2]== ' X ') gra[r2][c2]= ' * ';        DFS (R1,C1);        if (OK) printf ("yes\n");    else printf ("no\n"); } return 0;} /*4 6X... Xx... Xx.. X.. X...... X1 64 6*/

D-bad Luck Island

Main topic: Initial 3 species: R stone, s scissors, p cloth; the probability of any pair meeting is equal (regardless of the same kind of meeting); The probability of leaving only one species at the end;

Topic Analysis:

Probability knowledge, if present (stone, scissors, cloth): (i,j,k); then the probability is dp[I [j] [K] * (i * k + i * j + J * k) if the next step is to hang the stone (i-1,j,k));

others;

Code:

const int n=100007;double dp[110][110][110];int Main () {int r,s,p;                while (scanf ("%d%d%d", &r,&s,&p) ==3) {for (int. i=0;i<=r;i++) {for (int j=0;j<=s;j++) {                for (int k=0;k<=p;k++) {dp[i][j][k]=0.0;        }}} dp[r][s][p]=1.0;                    for (int i=r;i>=0;i--) {for (int. j=s;j>=0;j--) {for (int k=p;k>=0;k--) {                    if ((i==0&&j==0&&k==0)) continue; if (((i+1) *k+j*k+ (i+1) *j)//guarantee that this transfer can be established, the denominator is not 0; >0 (dp[i][j][k]+=1.0* (dp[i+1][j][k]*) i+1)/((*k) *                    k+j*k+ (i+1) *j);                    if (((j+1) *i+i*k+ (j+1) *k) >0 (dp[i][j][k]+=1.0* (dp[i][j+1][k]*) j+1)/((*i) j+1 (*i+i*k+) j+1); if ((i*j+i* (k+1) +j* (k+1) >0) dp[i][j][k]+=1.0* (dp[i][j][k+1]* (k+1) *j)/(i*j+i* (k+1                    ) +j* (k+1)); Cout<<dp[i][j][k]<<endL        }}} double a=0.0,b=0.0,c=0.0;        for (int i=1;i<=r;i++) a+=dp[i][0][0];        for (int j=1;j<=s;j++) b+=dp[0][j][0];        for (int k=1;k<=p;k++) c+=dp[0][0][k];    printf ("%.12f%.12f%.12f\n", a,b,c); } return 0;}

Cf!!!!!!!! When to div1!!!

CF 301 Div2