78 条题解
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1
Vincent (_Vincent) LV 5 @ 2021-11-13 21:51:01
#include<bits/stdc++.h> using namespace std; #define ull unsigned long long ull f[104][104],sum; int a[104]; inline int read() { register int x=0,f=0;register char ch=getchar(); while(ch<'0' || ch>'9')f|=ch=='-',ch=getchar(); while(ch>='0' && ch<='9')x=(x<<3)+(x<<1)+(ch^48),ch=getchar(); return f?-x:x; } signed main() { int n=read(),m=read(); for(int i=1;i<=n;i++)a[i]=read(),f[i][1]=1; for(int i=2;i<=n;i++) for(int j=i-1;j>=1;j--) for(int k=2;k<=i;k++) if(a[i]>=a[j]) f[i][k]=f[i][k]+f[j][k-1]; for(int i=1;i<=n;i++)sum+=f[i][m]; cout<<sum; return 0; } -
1
online_wlq LV 8 @ 2018-07-27 19:49:07
开longlong！！！！！！！！！！太重要了
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0
预感到数据很大，就立马开了unsigned long long，一次水过。
#include <iostream> #define ULL unsigned long long using namespace std; int n,m; ULL dp[101][21]={0};//dp[i][j] 以i结尾的长度为j的非递减字串种数 int a[101]={0}; int main() { cin>>n>>m; int i,j,k; for(i=1;i<=n;i++) { cin>>a[i]; } for(i=1;i<=n;i++) { for(j=1;j<=m;j++) { if(j==1) { dp[i][j]=1; } for(k=1;k<i;k++) { if(a[k]<=a[i]) { dp[i][j]+=dp[k][j-1]; } } } } ULL ans=0; for(i=1;i<=n;i++) { ans+=dp[i][m]; } cout<<ans<<endl; return 0; } -
0
Vijos解题排名**前1000**纪念！！
编译成功...
Free Pascal Compiler version 3.0.0 [2015/11/16] for i386
Copyright (c) 1993-2015 by Florian Klaempfl and others
Target OS: Win32 for i386
Compiling foo.pas
Linking foo.exe
24 lines compiled, 0.0 sec, 28176 bytes code, 1268 bytes data测试数据 #0: Accepted, time = 0 ms, mem = 8664 KiB, score = 5
测试数据 #1: Accepted, time = 15 ms, mem = 8664 KiB, score = 5
测试数据 #2: Accepted, time = 0 ms, mem = 8664 KiB, score = 5
测试数据 #3: Accepted, time = 0 ms, mem = 8664 KiB, score = 5
测试数据 #4: Accepted, time = 0 ms, mem = 8664 KiB, score = 5
测试数据 #5: Accepted, time = 15 ms, mem = 8664 KiB, score = 5
测试数据 #6: Accepted, time = 0 ms, mem = 8668 KiB, score = 5
测试数据 #7: Accepted, time = 0 ms, mem = 8664 KiB, score = 5
测试数据 #8: Accepted, time = 15 ms, mem = 8668 KiB, score = 5
测试数据 #9: Accepted, time = 0 ms, mem = 8664 KiB, score = 5
测试数据 #10: Accepted, time = 0 ms, mem = 8664 KiB, score = 5
测试数据 #11: Accepted, time = 0 ms, mem = 8664 KiB, score = 5
测试数据 #12: Accepted, time = 15 ms, mem = 8664 KiB, score = 5
测试数据 #13: Accepted, time = 15 ms, mem = 8668 KiB, score = 5
测试数据 #14: Accepted, time = 0 ms, mem = 8668 KiB, score = 5
测试数据 #15: Accepted, time = 15 ms, mem = 8664 KiB, score = 5
测试数据 #16: Accepted, time = 0 ms, mem = 8668 KiB, score = 5
测试数据 #17: Accepted, time = 0 ms, mem = 8668 KiB, score = 5
测试数据 #18: Accepted, time = 15 ms, mem = 8664 KiB, score = 5
测试数据 #19: Accepted, time = 15 ms, mem = 8668 KiB, score = 5
Accepted, time = 120 ms, mem = 8668 KiB, score = 100 -
0
tootal (hzq) LV 9 @ 2016-07-13 19:29:33
数据太弱了吧、、、三重循环都能A。。
设f[i][j]表示以A[i]结尾长度为j的不下降子序列个数
f[i][j]=sum(f[k][j-1]) (A[k]<=A[i])
~~~
#include <cstdio>
#include <cstdlib>
long long int n,m,ans,A[105],f[105][100];
int main(){
scanf("%lld%lld",&n,&m);
for(int i=1;i<=n;i++){
scanf("%lld",A+i);
}
for(int i=1;i<=n;i++)
f[i][1]=1;
for(int i=1;i<=n;i++){
for(int j=2;j<=m;j++){
if(j>i)break;
for(int k=1;k<i;k++)
if(A[k]<=A[i]) f[i][j]+=f[k][j-1];
}
}
for(int i=1;i<=n;i++){
ans+=f[i][m];
}
printf("%lld\n",ans);
return 0;
} -
0
！！！！！！！！！！！！！！！！注意开long long！！！！！！！！！！！！！
评测结果
编译成功测试数据 #0: Accepted, time = 0 ms, mem = 692 KiB, score = 5
测试数据 #1: Accepted, time = 0 ms, mem = 696 KiB, score = 5
测试数据 #2: Accepted, time = 0 ms, mem = 696 KiB, score = 5
测试数据 #3: Accepted, time = 0 ms, mem = 696 KiB, score = 5
测试数据 #4: Accepted, time = 0 ms, mem = 692 KiB, score = 5
测试数据 #5: Accepted, time = 0 ms, mem = 688 KiB, score = 5
测试数据 #6: Accepted, time = 0 ms, mem = 700 KiB, score = 5
测试数据 #7: Accepted, time = 0 ms, mem = 692 KiB, score = 5
测试数据 #8: Accepted, time = 0 ms, mem = 692 KiB, score = 5
测试数据 #9: Accepted, time = 0 ms, mem = 688 KiB, score = 5
测试数据 #10: Accepted, time = 0 ms, mem = 696 KiB, score = 5
测试数据 #11: Accepted, time = 0 ms, mem = 692 KiB, score = 5
测试数据 #12: Accepted, time = 0 ms, mem = 696 KiB, score = 5
测试数据 #13: Accepted, time = 0 ms, mem = 692 KiB, score = 5
测试数据 #14: Accepted, time = 0 ms, mem = 696 KiB, score = 5
测试数据 #15: Accepted, time = 0 ms, mem = 696 KiB, score = 5
测试数据 #16: Accepted, time = 0 ms, mem = 696 KiB, score = 5
测试数据 #17: Accepted, time = 0 ms, mem = 696 KiB, score = 5
测试数据 #18: Accepted, time = 0 ms, mem = 696 KiB, score = 5
测试数据 #19: Accepted, time = 15 ms, mem = 696 KiB, score = 5
Accepted, time = 15 ms, mem = 700 KiB, score = 100
代码
#include<cstdio>
#include<cstring>
int n,m,num[103],N=0;
long long ans=0,res[2][101][101];
int main(){
scanf ("%d%d",&n,&m);
memset(res[0],0,sizeof res[0]);
for (int i=1 ; i<=n ; i++) scanf("%d",&num[i]);
for (int z=1 ; z<=n ; z++) res[0][z][z]=1;
for (int z=2 ; z<=m ; z++){
N=1-N;
memset(res[N],0,sizeof res[N]);
for (int i=1 ; i<=n-z+1; i++)
for (int j=i+z-1; j<=n ; j++)
for (int k=i+1 ; k<=j-z+2; k++)
if ( num[ k ] >= num[ i ] ) res[N][i][j]+=res[1-N][k][j];
}
for (int i=1 ; i<=n-m+1; i++)
for (int j=i+m-1; j<=n ; j++) ans+=res[ N ][i][j];
printf("%I64d",ans);
} -
0
jiaoyunhao LV 10 @ 2012-08-12 17:09:11
手贱加了个高精
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0
SecretAgent LV 10 @ 2009-10-29 16:11:51
不仅是数组
ans也要用int64 -
0
编译通过...
├ 测试数据 01：答案正确... 0ms
├ 测试数据 02：答案正确... 0ms
├ 测试数据 03：答案正确... 0ms
├ 测试数据 04：答案正确... 0ms
├ 测试数据 05：答案正确... 0ms
├ 测试数据 06：答案正确... 0ms
├ 测试数据 07：答案正确... 0ms
├ 测试数据 08：答案正确... 0ms
├ 测试数据 09：答案正确... 0ms
├ 测试数据 10：答案正确... 0ms
├ 测试数据 11：答案正确... 0ms
├ 测试数据 12：答案正确... 0ms
├ 测试数据 13：答案正确... 0ms
├ 测试数据 14：答案正确... 0ms
├ 测试数据 15：答案正确... 0ms
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├ 测试数据 18：答案正确... 0ms
├ 测试数据 19：答案正确... 0ms
├ 测试数据 20：答案正确... 0ms
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Accepted 有效得分：100 有效耗时：0ms
哎...看了题解才知道用int64...我的AC率啊
f:=f+f[k,j-1];
初值f:=1;
最后把f累加.注意int64! -
0
violethill LV 10 @ 2009-10-18 15:03:29
第三个点longint会挂的 要开qword
题目那么长，全是废话 我看都没看 直接看输入格式就直接做了
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0
code-geass LV 9 @ 2009-09-16 18:03:03
为什么EXTENDED过不了第三组............INT64又可以..........why
Vincent (_Vincent)
LV 5
@ 2021-11-13 21:51:01
