codeforce 479E Riding in a Lift
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原题地址
题意
一个N层建筑,有一台电梯,一个人开始时位于a层,他觉得坐电梯上上下下K次,建筑的b层有一个实验室,实验室禁止进入,而且对于人每次坐电梯有限制条件
假如目前人在X层,希望坐电梯去Y层,需要满足
abs(X-Y)<abs(X-b)
问人这K次上上下下总共有多少种方案,答案对1e9+7取模
题解
DP啦,状态转移比较好搞吧,有两个问题,一个是严格大于的写法,一个是剩余系下减法要避免出现负数
#include<bits/stdc++.h>using namespace std;typedef long long LL;const int maxn=5000;const int H=1e9+7;int dp[maxn+5][maxn+5];int sum[maxn+5][maxn+5];int main(void){#ifdef exfreopen ("in.txt","r",stdin);#endifint n,a,b,k;scanf("%d%d%d%d",&n,&a,&b,&k);dp[0][a]=1;for (int i=a;i<=n;++i){sum[0][i]=1;}if (a<b){for (int i=1;i<=k;++i){for (int j=1;j<=b-1;++j){int h=ceil((b+j*1.0)/2-1);dp[i][j]=sum[i-1][h];if (j<=h) dp[i][j]=(dp[i][j]-dp[i-1][j]+H)%H;sum[i][j]=(sum[i][j-1]+dp[i][j])%H;}}printf("%d\n",sum[k][b-1]);#ifdef ex1cout<<dp[2][1]<<endl;cout<<dp[2][2]<<endl;cout<<dp[2][3]<<endl;#endif}else{for (int i=1;i<=k;++i){for (int j=b+1;j<=n;++j){int h=floor((b+j*1.0)/2+1);dp[i][j]=(sum[i-1][n]-sum[i-1][h-1]+H)%H;if (j>=h) dp[i][j]=(dp[i][j]-dp[i-1][j]+H)%H;sum[i][j]=(sum[i][j-1]+dp[i][j])%H;}}printf("%d\n",sum[k][n]-sum[k][b]);}}
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