1 条题解
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0
C++ :
#include <stdio.h> #include <string.h> #define MAX 10000000 #define MIN -10000000 int line[200],n,m,Min,Max,sum[200]; int f[200][20];//f[i][j]=第1-i个数分成j份,结果最大值 int g[200][20];//g[i][j]=第1-i个数分成j份,结果最小值 int min(int a,int b) { if(a>b) return b; return a; } int max(int a,int b) { if(a>b) return a; return b; } void dp(int a[]) { int i,j,k; for(i=1;i<=n;i++) sum[i]=sum[i-1]+a[i]; for(i=0;i<=n;i++) for(j=0;j<=m;j++) { f[i][j]=0; g[i][j]=-1u>>1; } for(i=1;i<=n;i++) { f[i][1]=g[i][1]=(sum[i]%10+10)%10; } f[0][0]=1; g[0][0]=1; for(j=2;j<=m;j++) { for(i=j;i<=n;i++) { for(k=j-1;k<i;k++) { { f[i][j]=max(f[i][j],f[k][j-1]*(((sum[i]-sum[k])%10+10)%10)); g[i][j]=min(g[i][j],g[k][j-1]*(((sum[i]-sum[k])%10+10)%10)); } } } } Max=max(Max,f[n][m]); Min=min(Min,g[n][m]); } int main() { //freopen("game.in","r",stdin); //freopen("game.out","w",stdout); int i,j,k; Max=0; Min=-1u>>1; scanf("%d%d",&n,&m); for(i=1;i<=n;i++) { scanf("%d",&line[i]); line[i+n]=line[i]; } for(i=0;i<n;i++) dp(line+i); printf("%d\n%d\n",Min,Max); return 0; }Pascal :
var a,atmp:array[0..50] of longint; f:array[0..9,0..50] of longint; n,m,i:longint; function g(a,b:longint):longint; var i:longint; begin g:=0; for i:=a to b do if atmp[i]>0 then g:=g+atmp[i] else g:=g+atmp[i]; g:=(g+100000) mod 10; end; procedure question_min; var t,i,j,k,tmp,min:longint; begin min:=maxlongint; for t:=1 to n do begin for i:=1 to n do atmp[i]:=a[(i+t-1) mod n]; atmp[n-t+1]:=a[n];f[0,1]:=atmp[1]; for i:=1 to n do f[0,i]:=g(1,i); for i:=1 to m-1 do for j:=i+1 to n do begin f[i,j]:=maxlongint; for k:=i to j-1 do begin tmp:=g(k+1,j); if f[i,j]>tmp*f[i-1,k] then f[i,j]:=tmp*f[i-1,k]; end; end; if f[m-1,n]<min then min:=f[m-1,n]; end; writeln(min); end; procedure question_max; var t,i,j,k,tmp,max:longint; begin max:=0; for t:=1 to n do begin for i:=1 to n do atmp[i]:=a[(i+t-1) mod n]; atmp[n-t+1]:=a[n];f[0,1]:=atmp[1]; for i:=1 to n do f[0,i]:=g(1,i); for i:=1 to m-1 do for j:=i+1 to n do begin f[i,j]:=0; for k:=i to j-1 do begin tmp:=g(k+1,j); if f[i,j]<tmp*f[i-1,k] then f[i,j]:=tmp*f[i-1,k]; end; end; if f[m-1,n]>max then max:=f[m-1,n]; end; writeln(max); end; begin {assign(input,'game.in');reset(input); assign(output,'game.out');rewrite(output);} readln(n,m); for i:=1 to n do readln(a[i]); question_min; question_max; {close(input); close(output);} end.
- 1
信息
- ID
- 1668
- 时间
- 1000ms
- 内存
- 128MiB
- 难度
- (无)
- 标签
- 递交数
- 0
- 已通过
- 0
- 上传者