CF1042C Array Product 分类讨论+贪心
Posted reverymoon
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考虑有无负数(负数的个数为奇视作“有”,否则为“无”)和有无零
无负数无零,全部合并即可
无负数有零,那么把零合并起来,删掉零
有负数无零,把最大的负数找出来,删掉,合并剩余的数
有负数有零,把零和最大的负数合并起来,删掉,合并剩余的数
注意如果只剩下一个数,不能删掉这唯一的一个数
#include <vector> #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> namespace remoon { #define ri register int #define ll long long #define tpr template <typename ra> #define rep(iu, st, ed) for(ri iu = st; iu <= ed; iu ++) #define drep(iu, ed, st) for(ri iu = ed; iu >= st; iu --) #define gc getchar inline int read() { int p = 0, w = 1; char c = gc(); while(c > ‘9‘ || c < ‘0‘) { if(c == ‘-‘) w = -1; c = gc(); } while(c >= ‘0‘ && c <= ‘9‘) p = p * 10 + c - ‘0‘, c = gc(); return p * w; } int wr[50], rw; #define pc(iw) putchar(iw) tpr inline void write(ra o, char c = ‘ ‘) { if(!o) pc(‘0‘); if(o < 0) o = -o, pc(‘-‘); while(o) wr[++ rw] = o % 10, o /= 10; while(rw) pc(wr[rw --] + ‘0‘); pc(c); } tpr inline void cmin(ra &a, ra b) { if(a > b) a = b; } tpr inline void cmax(ra &a, ra b) { if(a < b) a = b; } tpr inline bool ckmin(ra &a, ra b) { return (a > b) ? a = b, 1 : 0; } tpr inline bool ckmax(ra &a, ra b) { return (a < b) ? a = b, 1 : 0; } } using namespace std; using namespace remoon; #define sid 300050 int n, a[sid]; inline void Solve1() { rep(i, 2, n) printf("1 %d %d ", i, 1); } inline void Solve2() { int mip = -1e9 - 5, pos = 0; rep(i, 1, n) if(a[i] < 0 && ckmax(mip, a[i])) pos = i; int fir = -1; rep(i, 1, n) if(pos != i) { fir = i; break; } if(fir == -1) return; printf("2 %d ", pos); if(pos != 1) { rep(i, 2, n) if(pos != i) printf("%d %d %d ", 1, i, 1); } else rep(i, 3, n) printf("%d %d %d ", 1, i, 2); } inline void Solve3() { int lst = -1; rep(i, 1, n) if(a[i] == 0) { if(lst != -1) printf("%d %d %d ", 1, lst, i); lst = i; } int fir = -1; rep(i, 1, n) if(a[i] != 0) { fir = i; break; } if(fir == -1) return; printf("2 %d ", lst); rep(i, fir + 1, n) if(a[i] != 0) printf("%d %d %d ", 1, i, fir); } inline void Solve4() { int mip = -1e9 - 5, pos = 0; rep(i, 1, n) if(a[i] < 0 && ckmax(mip, a[i])) pos = i; int lst = -1; rep(i, 1, n) if(a[i] == 0 || pos == i) { if(lst != -1) printf("%d %d %d ", 1, lst, i); lst = i; } int fir = -1; rep(i, 1, n) if(a[i] != 0 && pos != i) { fir = i; break; } if(fir == -1) return; printf("2 %d ", lst); rep(i, fir + 1, n) if(a[i] != 0 && pos != i) printf("%d %d %d ", 1, i, fir); } int main() { n = read(); rep(i, 1, n) a[i] = read(); int neg = 0, zero = 0; rep(i, 1, n) if(a[i] < 0) neg ^= 1; else if(a[i] == 0) zero |= 1; if(!zero && !neg) Solve1(); else if(!zero && neg) Solve2(); else if(zero && !neg) Solve3(); else if(zero && neg) Solve4(); return 0; }
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