div1 250pt:
题意：。。。
解法：先假设空出来的位置是0,然后模拟一次看看是不是满足，如果不行的话，我们只需要关心最后栈顶的元素取值是不是受空白处的影响，于是还是模拟一下。
1 // BEGIN CUT HERE 2 3 // END CUT HERE 4 #line 5 "Suminator.cpp" 5 #include<cstdio> 6 #include<cstring> 7 #include<cstdlib> 8 #include<ctime> 9 #include<cmath> 10 #include<cassert> 11 #include<iostream> 12 #include<string> 13 #include<sstream> 14 #include<vector> 15 #include<map> 16 #include<set> 17 #include<queue> 18 #include<stack> 19 #include<algorithm> 20 using namespace std; 21 typedef long long ll; 22 typedef pair<int,int> pii; 23 const int N = 55; 24 ll stk[N]; 25 class Suminator 26 { 27 public: 28 int findMissing(vector <int> program, int wantedResult){ 29 //$CARETPOSITION$ 30 int top=0; 31 int n=program.size(); 32 for(int i=0;i<n;i++){ 33 if(program[i] == 0 || program[i] == -1){ 34 if(top >= 2){ 35 ll res = stk[top]+stk[top-1]; 36 stk[--top]=res; 37 // if(res > wantedResult)return -1; 38 }else if(top == 0){ 39 stk[++top] = 0; 40 } 41 }else{ 42 stk[++top] = program[i]; 43 } 44 } 45 if(stk[top] == wantedResult)return 0; 46 bool have[N];memset(have,0,sizeof(have)); 47 top = 0; 48 for(int i=0;i<n;i++){ 49 if(program[i] == 0){ 50 if(top >= 2){ 51 ll res = stk[top] + stk[top-1]; 52 bool ok = have[top] || have[top-1]; 53 stk[--top] = res; 54 // if(res > wantedResult)return -1; 55 have[top] = ok; 56 }else if(top == 0){ 57 stk[++top] = 0; 58 have[top] = 0; 59 } 60 }else if(program[i] == -1){ 61 stk[++top] = 0; 62 have[top] = 1; 63 }else{ 64 stk[++top] = program[i]; 65 have[top] = 0; 66 } 67 } 68 if(have[top] == 0)return -1; 69 else if(stk[top] >= wantedResult)return -1; 70 else return wantedResult - stk[top]; 71 72 } 73 74 // BEGIN CUT HERE 75 public: 76 void run_test(int Case) { if ((Case == -1) || (Case == 0)) test_case_0(); if ((Case == -1) || (Case == 1)) test_case_1(); if ((Case == -1) || (Case == 2)) test_case_2(); if ((Case == -1) || (Case == 3)) test_case_3(); if ((Case == -1) || (Case == 4)) test_case_4(); if ((Case == -1) || (Case == 5)) test_case_5(); } 77 private: 78 template <typename T> string print_array(const vector<T> &V) { ostringstream os; os << "{ "; for (typename vector<T>::const_iterator iter = V.begin(); iter != V.end(); ++iter) os << '"' << *iter << "","; os << " }"; return os.str(); } 79 void verify_case(int Case, const int &Expected, const int &Received) { cerr << "Test Case #" << Case << "..."; if (Expected == Received) cerr << "PASSED" << endl; else { cerr << "FAILED" << endl; cerr << " Expected: "" << Expected << '"' << endl; cerr << " Received: "" << Received << '"' << endl; } } 80 void test_case_0() { int Arr0[] = {7,-1,0}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 10; int Arg2 = 3; verify_case(0, Arg2, findMissing(Arg0, Arg1)); } 81 void test_case_1() { int Arr0[] = {100, 200, 300, 0, 100, -1}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 600; int Arg2 = 0; verify_case(1, Arg2, findMissing(Arg0, Arg1)); } 82 void test_case_2() { int Arr0[] = {-1, 7, 3, 0, 1, 2, 0, 0}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 13; int Arg2 = 0; verify_case(2, Arg2, findMissing(Arg0, Arg1)); } 83 void test_case_3() { int Arr0[] = {-1, 8, 4, 0, 1, 2, 0, 0}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 16; int Arg2 = -1; verify_case(3, Arg2, findMissing(Arg0, Arg1)); } 84 void test_case_4() { int Arr0[] = {1000000000, 1000000000, 1000000000, 1000000000, -1, 0, 0, 0, 0}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 1000000000; int Arg2 = -1; verify_case(4, Arg2, findMissing(Arg0, Arg1)); } 85 void test_case_5() { int Arr0[] = {7, -1, 3, 0}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 3; int Arg2 = -1; verify_case(5, Arg2, findMissing(Arg0, Arg1)); } 86 87 // END CUT HERE 88 89 }; 90 // BEGIN CUT HERE 91 int main(){ 92 Suminator ___test; 93 ___test.run_test(-1); 94 return 0; 95 } 96 // END CUT HERE
div1 500pt
有点复杂的dp，有待实现