題意：較大的容量減較小的容量，較小的容量翻倍。問操作幾回其中一個空。

開始用set判重，重復就不可行。不過狀態最多有2e18種。不僅爆內存，還超時。然后找規律。發現只有比例為1:1,1:3,1:7,3:5,1:15,3:13,5:11,7:9......這樣才行。也就是化簡以后相加是2^k。

#pragma comment(linker,"/STACK:1024000000,1024000000") #include <iostream> #include <cstdio> #include <cmath> #include <algorithm> #include <vector> #include <iomanip> #include <cstring> #include <map> #include <queue> #include <set> #include <cassert> #include <list> #define mkp make_pair using namespace std; const double EPS=1e-8; const int SZ=1e2+10,INF=0x7FFFFFFF; const long long mod=19999997; typedef long long lon;int gcd(int x,int y) {if(x==0)return y;if(y==0)return x;if(x<y)swap(x,y);int res=0;for(;;){int rem=x%y;if(rem==0)return y;x=y;y=rem;} }int cnt(int x) {int res=0;for(;x;x-=x&-x,++res);return res; }int chk(int x) {int num=cnt(x);if(num!=1)return -1;else{for(int i=0;i<32;++i){if((1<<i)&x)return i;}} }int main() {std::ios::sync_with_stdio(0);//freopen("d:\\1.txt","r",stdin);int casenum;//cin>>casenum;//scanf("%d",&casenum);//for(int time=1;time<=casenum;++time)//for(int time=1;cin>>n;++time) {int n,m;cin>>n>>m;int d=gcd(n,m);//cout<<d<<endl;n/=d,m/=d; if(n==0||m==0)cout<<0<<endl;else{cout<<chk(n+m)<<endl;}}return 0; }

超內存的：

#include <iostream> #include <cstdio> #include <cmath> #include <algorithm> #include <vector> #include <iomanip> #include <cstring> #include <map> #include <queue> #include <set> #include <cassert> #include <list> using namespace std; const double EPS=1e-8; const int SZ=1e2+10,INF=0x7FFFFFFF; const long long mod=19999997; typedef long long lon; int n,sum[9],arr[9];bool chk(int x,int t) {int res=0;if(x/n&&arr[x]<arr[x-n])++res;if(x%n&&arr[x]<arr[x-1])++res;if(x/n!=n-1){if(t){if(arr[x]<arr[x+n])++res;}else ++res;}if(x%n!=n-1){if(t){if(arr[x]<arr[x+1])++res;}else ++res;}return t?res==sum[x]:res>=sum[x]; }bool dfs(int x) {//cout<<x<<endl;if(x==n*n){ // for(int i=0;i<n*n;++i) // { // cout<<arr[i]<<" "; // }cout<<endl;return 1;}for(int i=1;i<10;++i){arr[x]=i;if(x%n&&chk(x-1,0||((x-1)/n==n-1))==0)continue;if(x/n&&chk(x-n,1)==0)continue;if(x==n*n-1&&chk(x,1)==0)continue;//cout<<x<<" "<<i<<endl;if(dfs(x+1))return 1;//else cout<<x<<" "<<i<<" fail"<<endl; }arr[x]=0;return 0; }int main() {std::ios::sync_with_stdio(0);//freopen("d:\\1.txt","r",stdin);int casenum;//cin>>casenum;//scanf("%d",&casenum);//for(int time=1;time<=casenum;++time)//for(int time=1;cin>>n;++time) {cin>>n;for(int i=0;i<n*n;++i){cin>>sum[i];}if(n==1){if(sum[0]==0)cout<<1<<endl;else cout<<"NO SOLUTION"<<endl;}else if(dfs(0)){for(int i=0;i<n*n;++i){if(i%n)cout<<" ";cout<<arr[i];if(i%n==n-1)cout<<endl;}}else{cout<<"NO SOLUTION"<<endl;}}return 0; }

超時的：

#pragma comment(linker,"/STACK:1024000000,1024000000") #include <iostream> #include <cstdio> #include <cmath> #include <algorithm> #include <vector> #include <iomanip> #include <cstring> #include <map> #include <queue> #include <set> #include <cassert> #include <list> #define mkp make_pair using namespace std; const double EPS=1e-8; const int SZ=1e2+10,INF=0x7FFFFFFF; const long long mod=19999997; typedef long long lon;int work(int x,int y,int x1,int y1,int s) {if(x<y)swap(x,y);if(x1<y1)swap(x1,y1);//cout<<x<<" "<<y<<" "<<x1<<" "<<y1<<endl;if(s!=1&&x==x1&&y==y1)return -1;x-=y;y*=2;x1-=y1;y1*=2;if(x<y)swap(x,y);if(x1<y1)swap(x1,y1);//if(s!=1&&x==x1&&y==y1)return -1;x1-=y1;y1*=2;if(x==0||y==0)return s;return work(x,y,x1,y1,s+1); }int main() {std::ios::sync_with_stdio(0);//freopen("d:\\1.txt","r",stdin);int casenum;//cin>>casenum;//scanf("%d",&casenum);//for(int time=1;time<=casenum;++time)//for(int time=1;cin>>n;++time) {int n,m;cin>>n>>m;if(n==0||m==0)cout<<0<<endl;else if((n&1)^(m&1))cout<<-1<<endl;else cout<<work(n,m,n,m,1)<<endl;}return 0; }

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轉載于:https://www.cnblogs.com/gaudar/p/9764809.html

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