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A. Dreamoon and Stairs
题解:
首先写出尽可能2多的步数，然后判断能否%m,不能就加上最小的数使其能%m就行了
代码:
#include<bits/stdc++.h> using namespace std; #define pb push_back #define mp make_pair #define se second #define fs first #define LL long long #define CLR(x) memset(x,0,sizeof x) #define MC(x,y) memcpy(x,y,sizeof(x)) #define SZ(x) ((int)(x).size()) #define FOR(it,c) for(__typeof((c).begin()) it=(c).begin();it!=(c).end();it++) #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 typedef pair<int,int> P; const double eps=1e-9; const int N=1e5+10; const int M=1e3+10; const int mod=1e9+7; const int INF=1e9+10; int n,m; int main(){ int n,m; cin>>n>>m; if(n<m) cout<<-1<<endl; else{ int ans=n/2+(n%2==0?0:1); if(ans%m==0) cout<<ans<<endl; else{ for(int i=1;i<m;i++){ if((ans+i)%m==0){ cout<<ans+i<<endl; break; } } } } }
B.Dreamoon and WiFi
题解:
简单dp,递推搜索一下就行了
代码:
#include<bits/stdc++.h> using namespace std; #define pb push_back #define mp make_pair #define se second #define fs first #define LL long long #define CLR(x) memset(x,0,sizeof x) #define MC(x,y) memcpy(x,y,sizeof(x)) #define SZ(x) ((int)(x).size()) #define FOR(it,c) for(__typeof((c).begin()) it=(c).begin();it!=(c).end();it++) #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 typedef pair<int,int> P; const double eps=1e-9; const int N=1e5+10; const int M=1e3+10; const int mod=1e9+7; const int INF=1e9+10; string s1,s2; int cnt1,cnt,len; void DP(int pos,int c){ if(pos==len){ if(c==cnt1) cnt++; return; } if(s2[pos]==‘+‘) DP(pos+1,c+1); if(s2[pos]==‘-‘) DP(pos+1,c-1); if(s2[pos]==‘?‘){ DP(pos+1,c-1); DP(pos+1,c+1); } } int main(){ cin>>s1>>s2; cnt1=0,cnt=0; len=s1.length(); for(int i=0;i<len;i++) if(s1[i]==‘+‘) cnt1++;else cnt1--; int sum=1; for(int i=0;i<len;i++) if(s2[i]==‘?‘) sum*=2; DP(0,0); cout<<setprecision(15)<<(double)cnt/(double)sum<<endl; return 0; }
C.Dreamoon and Sums
题解:
注意到x/b=x/b*b+x%b就行。然后注意每个计算位置都要计算取mod.....
代码:
#include<bits/stdc++.h> using namespace std; #define pb push_back #define mp make_pair #define se second #define fs first #define LL long long #define CLR(x) memset(x,0,sizeof x) #define MC(x,y) memcpy(x,y,sizeof(x)) #define SZ(x) ((int)(x).size()) #define FOR(it,c) for(__typeof((c).begin()) it=(c).begin();it!=(c).end();it++) #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 typedef pair<int,int> P; const double eps=1e-9; const int N=1e5+10; const int M=1e3+10; const LL mod=1000000007; const int INF=1e9+10; LL a,b; int main(){ cin>>a>>b; LL k=((1LL+a)*a/2%mod*b%mod+a)%mod; LL sum=0; for(LL i=1;i<b;i++){ sum+=i*k; sum%=mod; } cout<<sum<<endl; }
D. Dreamoon and Sets
题解:
规律题，不多说
代码:
#include <iostream> using namespace std; int main(){ int n,k; cin>>k>>n; cout<<(6*k-1)*n<<endl; for(int i=0;i<k;i++)cout<<n*(6*i+1)<<" "<<n*(6*i+2)<<" "<<n*(6*i+3)<<" "<<n*(6*i+5)<<endl; }
E. Dreamoon and Strings
题解:
暴力处理每个位置，从后面到前面的第一个模板的位置，然后dp处理
参考博客http://www.cnblogs.com/qscqesze/p/5794709.html
代码:
//Coding by qscqesze #include<bits/stdc++.h> using namespace std; const int maxn = 2005; char s1[maxn],s2[maxn]; int dp[maxn][maxn]; int len1,len2; int solve(int x) { if(x<len2)return maxn; int a=x,b=len2,tmp=0; while(a&&b) { if(s1[a]==s2[b])a--,b--; else tmp++,a--; } if(b==0)return tmp; else return maxn; } int main() { scanf("%s%s",s1+1,s2+1); len1 = strlen(s1+1); len2 = strlen(s2+1); for(int i=0;i<=len1;i++) for(int j=0;j<=len1;j++) if(j>i)dp[i][j]=-3000; for(int i=1;i<=len1;i++) { int x=solve(i); for(int k=0;k<=len1;k++) dp[i][k]=max(dp[i][k],dp[i-1][k]); for(int k=0;k<=len1;k++)if(x<=k) dp[i][k]=max(dp[i][k],dp[i-x-len2][k-x]+1); } for(int i=0;i<=len1;i++) printf("%d ",dp[len1][i]); printf("\n"); }
Codeforces Round #272 (Div. 2) ABCDE
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原文地址：http://www.cnblogs.com/byene/p/5857820.html
