Codeforces Round #318 (Div. 2) (A,B,C,D)

Question A:
Title Address: Bear and elections
Test instructions: The minimum number of transformations can make the first digit larger than all subsequent digits.
Idea: The number of the following n-1, let the first and last number comparison, and then increase or decrease. Knowing that the first number is greater than the last number

#include <stdio.h>#include <math.h>#include <string.h>#include <stdlib.h>#include <iostream>#include <sstream>#include <algorithm>#include <set>#include <queue>#include <stack>#include <map>#include <bitset>#pragma COMMENT (linker, "/stack:102400000,102400000")using namespace STD;typedef Long LongLL;Const intinf=0x3f3f3f3f;Const DoublePi=ACOs(-1.0);Const Doubleesp=1e-6;using namespace STD;inta[ the];intMain () {intNintCnt while(~scanf("%d", &n)) { for(intI=0; i<n;i++)scanf("%d", &a[i]); Cnt=0; while(1) {sort (A +1, a+n);if(a[0]<=a[n-1]) {a[0]++; a[n-1]--;            cnt++; }Else                 Break; }printf("%d\n", CNT); }return 0;}

Question B:
Title Address: Bear and Three Musketeers
Test instructions: give you a figure and then determine if you can make a triangle, if you can, output the smallest number of points triangle, if not output-1

#include <stdio.h>#include <math.h>#include <string.h>#include <stdlib.h>#include <iostream>#include <sstream>#include <algorithm>#include <set>#include <queue>#include <stack>#include <map>#include <bitset>#pragma COMMENT (linker, "/stack:102400000,102400000")using namespace STD;typedef Long LongLL;Const intinf=0x3f3f3f3f;Const DoublePi=ACOs(-1.0);Const Doubleesp=1e-6;using namespace STD;Const intmaxn=4010;intVIS[MAXN][MAXN];intDEG[MAXN];intMain () {intN,m;intx, y; while(~scanf("%d%d", &n,&m)) {memset(Vis,0,sizeof(VIS));memset(DEG,0,sizeof(deg)); while(m--) {scanf("%d%d", &x,&y);            deg[x]++;            deg[y]++; vis[x][y]=vis[y][x]=1; }intMin=inf; for(intI=1; i<=n;i++) { for(intj=i+1; j<=n;j++) {if(Vis[i][j]) { for(intk=j+1; k<=n;k++) {if(Vis[k][j]&&vis[k][i]) {Min=min (min,deg[i]+deg[j]+deg[k]-6); }                    }                }            }        }if(Min==inf)puts("-1");Else            printf("%d\n", Min); }return 0;}

Question C:
Title Address: Bear and Poker
Test instructions: An array of length n, each of which can become 2 or 3 times times its own, asking if you can change the n number to a number.
Idea: Divide each number by 2 and 3 until there is no quality factor of 2 and 3, and then add to set, if only one number is left in the set then you can, otherwise you cannot.

#include <stdio.h>#include <math.h>#include <string.h>#include <stdlib.h>#include <iostream>#include <sstream>#include <algorithm>#include <set>#include <queue>#include <stack>#include <map>#include <bitset>#pragma COMMENT (linker, "/stack:102400000,102400000")using namespace STD;typedef Long LongLL;Const intinf=0x3f3f3f3f;Const DoublePi=ACOs(-1.0);Const Doubleesp=1e-6;using namespace STD;Const intmaxn=1e5+Ten; Set<int >QintMain () {intN,x; while(~scanf("%d", &n)) { while(n--) {scanf("%d", &x); while(%2==0) x/=2; while(%3==0) x/=3;        Q.insert (x); }if(q.size () = =1)puts("Yes");Else            puts("No"); }return 0;}

Question d:
Title Address: Bear and Blocks
Test instructions: Give a sequence of rows, each time the outer block is removed, ask the total minimum number of steps.
Idea: To find the minimum number of steps for each one from the back, to find the minimum number of steps from the rear to the next, and then to find the minimum number of steps for each, then the two compare to find the current minimum. And then find all the biggest inside (because to meet all the circumstances)

#include <stdio.h>#include <math.h>#include <string.h>#include <stdlib.h>#include <iostream>#include <sstream>#include <algorithm>#include <set>#include <queue>#include <stack>#include <map>#include <bitset>#pragma COMMENT (linker, "/stack:102400000,102400000")using namespace STD;typedef Long LongLL;Const intinf=0x3f3f3f3f;Const DoublePi=ACOs(-1.0);Const Doubleesp=1e-6;using namespace STD;Const intmaxn=1e5+Ten;intH[MAXN];intDP1[MAXN],DP2[MAXN];intRES[MAXN];intMain () {intN while(~scanf("%d", &n)) { for(intI=0; i<n;i++)scanf("%d", &h[i]); dp1[0]=1; for(intI=1; i<n;i++) Dp1[i]=min (dp1[i-1]+1, H[i]); dp2[n-1]=1; for(inti=n-2; i>=0; i--) Dp2[i]=min (dp2[i+1]+1, H[i]); for(intI=0; i<n;i++) {res[i]=min (dp1[i],dp2[i]); } sort (res,res+n);printf("%d\n", res[n-1]); }}

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Codeforces Round #318 (Div. 2) (A,B,C,D)