USACO 2.2

PROB Preface Numbering [ANALYSIS] ---- enumeration again, so I have to trust the strength of enumeration.

It should be the role of estimating the amount of data, which is an order of 1000, But it is strange that I still want to deploy by bit DP ....

 

int n;int sum_ge[7];//idx 1 - 3void calculate(int x,int idx){switch(x) {case 0:break;case 1:sum_ge[2*(idx-1)]++;break;case 2:sum_ge[2*(idx-1)]+=2;break;case 3:sum_ge[2*(idx-1)]+=3;break;case 4:sum_ge[2*(idx-1)]+=1;sum_ge[2*(idx-1)+1]++;break;case 5:sum_ge[2*(idx-1)+1]++;break;case 6:sum_ge[2*(idx-1)]+=1;sum_ge[2*(idx-1)+1]++;break;case 7:sum_ge[2*(idx-1)]+=2;sum_ge[2*(idx-1)+1]++;break;case 8:sum_ge[2*(idx-1)]+=3;sum_ge[2*(idx-1)+1]++;break;case 9:sum_ge[2*(idx-1)]+=1;sum_ge[2*(idx-1)+2]++;break;}}int main(){FOPENTIFOPENTOSET(sum_ge,0);SCF(n);FOR(i,1,n) {int t=1,x=i;while(x) {calculate(x%10,t++);x /= 10;}}if(sum_ge[0]) printf("I %d\n",sum_ge[0]);if(sum_ge[1]) printf("V %d\n",sum_ge[1]);if(sum_ge[2]) printf("X %d\n",sum_ge[2]);if(sum_ge[3]) printf("L %d\n",sum_ge[3]);if(sum_ge[4]) printf("C %d\n",sum_ge[4]);if(sum_ge[5]) printf("D %d\n",sum_ge[5]);if(sum_ge[6]) printf("M %d\n",sum_ge[6]);}

USER: Rain M [m3324631]TASK: prefaceLANG: C++Compiling...Compile: OKExecuting...   Test 1: TEST OK [0.000 secs, 3048 KB]   Test 2: TEST OK [0.000 secs, 3048 KB]   Test 3: TEST OK [0.000 secs, 3048 KB]   Test 4: TEST OK [0.000 secs, 3048 KB]   Test 5: TEST OK [0.000 secs, 3048 KB]   Test 6: TEST OK [0.000 secs, 3048 KB]   Test 7: TEST OK [0.000 secs, 3048 KB]   Test 8: TEST OK [0.000 secs, 3048 KB]All tests OK.YOUR PROGRAM ('preface') WORKED FIRST TIME!  That's fantastic-- and a rare thing.  Please accept these special automatedcongratulations.Here are the test data inputs:------- test 1 ----1------- test 2 ----20------- test 3 ----100------- test 4 ----500------- test 5 ----1000------- test 6 ----2974------- test 7 ----3213------- test 8 ----3499Keep up the good work!Thanks for your submission!

 

PROB Subset Sums [ANALYSIS] ---- obviously it's a backpack, It's a paste ..

// Solution 1 summation, DP // solution 1 // 1: summation, to see if it is an integer // 2: half of the sum of DP, how many components are there? // This is not repeated. After a long time, I finally realized that // This is the general backpack int n, sumhalf, ans; LLong anst [MAXN]; int main () {fopentifopento scf (n); int sum = (n + 1) * n)/2; if (sum & 0x1) puts ("0"); else {anst [0] = 1; for (int I = 1; I <= n; I ++) {for (int j = sum/2; j> = I; j --) {anst [j] + = anst [j-I] ;}} printf ("% lld \ n", anst [sum/2]/2 );}}

 

USER: Rain M [m3324631]TASK: subsetLANG: C++Compiling...Compile: OKExecuting...   Test 1: TEST OK [0.000 secs, 3052 KB]   Test 2: TEST OK [0.000 secs, 3052 KB]   Test 3: TEST OK [0.000 secs, 3052 KB]   Test 4: TEST OK [0.000 secs, 3052 KB]   Test 5: TEST OK [0.000 secs, 3052 KB]   Test 6: TEST OK [0.000 secs, 3052 KB]   Test 7: TEST OK [0.000 secs, 3052 KB]All tests OK.YOUR PROGRAM ('subset') WORKED FIRST TIME!  That's fantastic-- and a rare thing.  Please accept these special automatedcongratulations.Here are the test data inputs:------- test 1 ----7------- test 2 ----15------- test 3 ----24------- test 4 ----31------- test 5 ----36------- test 6 ----39------- test 7 ----37Keep up the good work!Thanks for your submission!

 

PROB Runaround Numbers [ANALYSIS] ---- enumeration. I have read the question several times.

There are many restrictions. Each number cannot be repeated. There cannot be 0. This number can only be a ring!

unsigned int n;bool flag;bool Ans[MAXN];bool check(unsigned int x){char ss[MAXN];SET(Ans,false);int chee = 0;sprintf(ss,"%u",x);int len = strlen(ss);F(i,len) {if(chee & (1<<(ss[i]-'0'))) return false;chee |= (1<<(ss[i]-'0'));}for(int i = 0;;) {int k = ss[i] - '0',che = i;if(k == 0) return false;if(Ans[i]) {if(i == 0) flag = 1;else return false;break;}else Ans[i] = true;while(k--) {if(++i >= len) i = 0;}if(ss[i] == ss[che]) return false;}F(i,len) {if(!Ans[i]) return false;}if(flag)return true;}int main(){        FOPENTI        FOPENTOscanf("%u",&n);while(!check(++n));printf("%u\n",n);}

 

USER: Rain M [m3324631]TASK: runroundLANG: C++Compiling...Compile: OKExecuting...   Test 1: TEST OK [0.000 secs, 3048 KB]   Test 2: TEST OK [0.011 secs, 3048 KB]   Test 3: TEST OK [0.000 secs, 3048 KB]   Test 4: TEST OK [0.011 secs, 3048 KB]   Test 5: TEST OK [0.108 secs, 3048 KB]   Test 6: TEST OK [0.065 secs, 3048 KB]   Test 7: TEST OK [0.162 secs, 3048 KB]All tests OK.YOUR PROGRAM ('runround') WORKED FIRST TIME!  That's fantastic-- and a rare thing.  Please accept these special automatedcongratulations.Here are the test data inputs:------- test 1 ----99------- test 2 ----111110------- test 3 ----134259------- test 4 ----348761------- test 5 ----1000000------- test 6 ----5000000------- test 7 ----9000000Keep up the good work!Thanks for your submission!

 

PROB Party Lamps [ANALYSIS] ---- Simulation

struct str_hash {        size_t operator()(const string& str) const        {                return __stl_hash_string(str.c_str());}};int n,stepgoal;char goal[105],status[105];map<string,int> has;vector<string> vec;void set_bit(int x,int bol){        goal[x-1] = bol + '0';}bool findans(){        F(i,n) {                if(goal[i] == '3') continue;                if(status[i] != goal[i]) return 0;        }        return 1;}void dfs(int depth){        if(findans()) {        vec.push_back(status);        }        if(depth == stepgoal) {        return;        }        //1        F(i,n) status[i] ='1'-status[i]+'0';        if(!has[status])        has[status] = 1,dfs(depth + 1);        F(i,n) status[i] ='1'-status[i]+'0';        //2        for(int i = 0;i<n;i+=2) {        status[i] ='1'-status[i]+'0';        }        if(!has[status])        has[status] = 1,dfs(depth + 1);        for(int i = 0;i<n;i+=2) {        status[i] ='1'-status[i]+'0';        }        //3        for(int i = 1;i<n;i+=2) {        status[i] ='1'-status[i]+'0';        }        if(!has[status])        has[status] = 1,dfs(depth + 1);        for(int i = 1;i<n;i+=2) {        status[i] ='1'-status[i]+'0';        }        //4        for(int i = 0;i<n;i+=3) {        status[i] ='1'-status[i]+'0';        }        if(!has[status])        has[status] = 1,dfs(depth + 1);        for(int i = 0;i<n;i+=3) {        status[i] ='1'-status[i]+'0';        }}int main(){FOPENTIFOPENTO        SCFD(n,stepgoal);        SET(goal,'\0');        F(i,n) goal[i]='3';        int a;        SCF(a);        while(a != -1) {                set_bit(a,1);                SCF(a);        }        SCF(a);        while(a != -1) {                set_bit(a,0);                SCF(a);        }        SET(status,'\0');        F(i,n)status[i] = '1';        has.clear();        has[status] = 1;        dfs(0);        sort(vec.begin(),vec.end());        F(i,vec.size()) {                cout<<vec[i]<<endl;        }        if(vec.size() == 0) {puts("IMPOSSIBLE");        }}

 

At first, it was time to forget that there was no answer. WA once

USER: Rain M [m3324631]TASK: lampsLANG: C++Compiling...Compile: OKExecuting...   Test 1: TEST OK [0.000 secs, 3196 KB]   Test 2: TEST OK [0.000 secs, 3196 KB]   Test 3: TEST OK [0.000 secs, 3196 KB]   Test 4: TEST OK [0.000 secs, 3196 KB]   Test 5: TEST OK [0.000 secs, 3196 KB]   Test 6: TEST OK [0.000 secs, 3196 KB]   Test 7: TEST OK [0.000 secs, 3196 KB]   Test 8: TEST OK [0.000 secs, 3196 KB]All tests OK.Your program ('lamps') produced all correct answers!  This is yoursubmission #2 for this problem.  Congratulations!Here are the test data inputs:------- test 1 ----100-1-1------- test 2 ----100-11 -1------- test 3 ----203-11 3 5 -1------- test 4 ----501001 -1-1------- test 5 ----75250-1-1------- test 6 ----10083941 7 13 19 25 31 37 43 49 55 -164 -1------- test 7 ----100200031 86 23 -142 -1------- test 8 ----1008950-1-1Keep up the good work!Thanks for your submission!

 

Good luck