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Codeforces 106 DIV2 ACD
B表示完全看不懂。。就不弄了。。
E字符串先不管了。到时候系统学下字符串再处理
A
#include <map>#include <set>#include <list>#include <cmath>#include <ctime>#include <deque>#include <stack>#include <queue>#include <cctype>#include <cstdio>#include <string>#include <vector>#include <climits>#include <cstdlib>#include <cstring>#include <iostream>#include <algorithm>#define LL long long#define PI 3.1415926535897932626using namespace std;int gcd(int a, int b) {return a % b == 0 ? b : gcd(b, a % b);}int src[13];int main(){ int K; scanf("%d",&K); for (int i = 0; i < 12; i++) scanf("%d",&src[i]); sort(src,src+12); int ans = 0 ,sum = 0, cas = 11; while (true) { if (sum >= K) break; if (ans == 13) break; sum += src[cas]; cas--; ans++; } if (ans == 13) puts("-1"); else printf("%d\n",ans); return 0;}
C
#include <map>#include <set>#include <list>#include <cmath>#include <ctime>#include <deque>#include <stack>#include <queue>#include <cctype>#include <cstdio>#include <string>#include <vector>#include <climits>#include <cstdlib>#include <cstring>#include <iostream>#include <algorithm>#define LL long long#define PI 3.1415926535897932626using namespace std;int gcd(int a, int b) {return a % b == 0 ? b : gcd(b, a % b);}#define MAXN 100005int N;struct node{ int id; int val; friend bool operator < (const node & a,const node &b) { return a.val < b.val; }}src[MAXN];LL sum[MAXN];node stl[MAXN],str[MAXN];int main(){ scanf("%d",&N); sum[0] = 0; for (int i = 1; i <= N; i++) { scanf("%d",&src[i].val); src[i].id = i ; sum[i] = sum[i - 1] + src[i].val; } sort(src + 1, src + 1 + N); int topl = 1 ,topr = 1; if (N % 2 == 0) { for (int i = 1; i <= N; i++) { if (i % 2 == 0) stl[topl++] =src[i]; else str[topr++] = src[i]; } printf("%d\n",N / 2); for (int i = 1; i < topl; i++) printf("%d ",stl[i].id); putchar(‘\n‘); printf("%d\n",N / 2); for (int i = 1; i < topr; i++) printf("%d ",str[i].id); putchar(‘\n‘); } else { for (int i = 1; i <= N; i += 2) stl[topl++] =src[i]; for (int i = 2; i <= N; i += 2) str[topr++] = src[i]; printf("%d\n",N / 2 + 1); for (int i = 1; i < topl; i++ ) printf("%d ",stl[i].id); putchar(‘\n‘); printf("%d\n", N / 2); for (int i = 1; i < topr; i++) printf("%d ",str[i].id); putchar(‘\n‘); } return 0;}
D
#include <map>#include <set>#include <list>#include <cmath>#include <ctime>#include <deque>#include <stack>#include <queue>#include <cctype>#include <cstdio>#include <string>#include <vector>#include <climits>#include <cstdlib>#include <cstring>#include <iostream>#include <algorithm>#define LL long long#define PI 3.1415926535897932626using namespace std;int gcd(int a, int b) {return a % b == 0 ? b : gcd(b, a % b);}#define MOD 1000000007char input[720];struct node{ char res; int id;}src[720];int G[720];node sta[720];LL dp[720][720][4][4];void calcu(int l ,int r){ if (l >= r) return ; if (l + 1 == r) { dp[l][r][0][1] = 1; dp[l][r][1][0] = 1; dp[l][r][2][0] = 1; dp[l][r][0][2] = 1; } if (G[l] == r) { calcu(l + 1,r - 1); for (int i = 0 ; i < 3; i++) for (int j = 0; j < 3; j++) { if (j != 1) dp[l][r][0][1] = (dp[l][r][0][1] + dp[l + 1][r - 1][i][j]) % MOD; if (j != 2) dp[l][r][0][2] = (dp[l][r][0][2] + dp[l + 1][r - 1][i][j]) % MOD; if (i != 1) dp[l][r][1][0] = (dp[l][r][1][0] + dp[l + 1][r - 1][i][j]) % MOD; if (i != 2) dp[l][r][2][0] = (dp[l][r][2][0] + dp[l + 1][r - 1][i][j]) % MOD; } return ; } else { int t = G[l]; calcu(l,t); calcu(t + 1, r); for (int i = 0 ; i < 3; i++) for (int j = 0 ; j < 3; j++) for (int m = 0; m < 3; m++) for (int n = 0 ; n < 3; n++) { if (!( (m == 1 && n == 1) || (m == 2 && n == 2))) dp[l][r][i][j] = (dp[l][r][i][j] + (dp[l][t][i][m] * dp[t + 1][r][n][j]) % MOD) % MOD; } }}int main(){ while (scanf("%s",input + 1) != EOF) { int len = strlen(input + 1); int top = 1; for (int i = 1; i <= len; i++) { if (input[i] == ‘(‘) { node tmp; tmp.id = i; tmp.res = input[i]; sta[top++] = tmp; } else { if (sta[top - 1].res == ‘(‘) { G[i] = sta[top - 1].id; G[sta[top - 1].id] = i; top--; } else { node tmp; tmp.id = i; tmp.res = input[i]; sta[top++] = tmp; } } } memset(dp,0,sizeof(dp)); calcu(1,len); LL ans = 0; for (int i = 0 ; i < 3; i++) for (int j = 0 ; j < 3; j++) ans = (ans + dp[1][len][i][j]) % MOD; printf("%I64d\n",ans % MOD); } return 0;}
Codeforces 106 DIV2 ACD
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