[cqoi2013]二进制a+b
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这道题蛮有意思的..
可以考虑DP，这样来设计状态：
f(ma,mb,mc,n,q)表示a用了ma个1，b用了mb个1，c用了mc个1，当前到第n位，q∈{0,1}表示是否有进位
然后只需枚举当前位a取0，b取0、a取1，b取1、a取1，b取0这三种情况，根据q来判断mc需要增加多少和答案需要增加多少
最后的答案就是f(na,nb,nc,len,0),na是a中1的个数，nb是b中1的个数，nc是c中1的个数,len是最长的数的长度
这里q一定要取0是因为，当我们得到一个状态时，若q等于1，那么n是比实际上的长度少1的，因为当前位（也就是需要枚举的这一位，也就是n+1位）有一个1.
至于转移么，我写了个SPFA转移（而且写慢了..懒得改了..)，应该有更好的转移方式（SPFA转移写了整整4K..)
那么完整代码如下:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232 233 234 235 236 237 238 239 240 241 242 243 244 245 246 247 248 | #include <stdio.h>#include <queue>#include <string.h> typedef long long Long; Long a = 0;Long b = 0;Long c = 0;int na = 0;int nb = 0;int nc = 0;int len = 0; int get_num(Long n){ Long m = (1LL << 60); int r = 0; while(m) { if(m & n) r++; m >>= 1; } return r;} int get_len(Long n){ bool flag = false; Long m = (1LL << 60); int r = 0; while(m) { if(m & n) flag = true; if(flag) r++; m >>= 1; } return r; } int MIN(int a,int b){ return a > b ? b : a;} Long MAX(Long a,Long b){ return a > b ? a : b;} const int N = 30; void upmin(Long & m,Long n){ if(n < 0) return; if(m > n || m < 0) m = n;} const Long INF = (1LL << 60); Long pow(Long a,int b){ Long m = (1LL<<60); Long r = 1LL; while(m) { r *= r; if(m & b) r *= a; m >>= 1; } return r;} struct node{ int ma; int mb; int mc; int q; int n; Long val; node() {} node(int MA,int MB,int MC,int NN,Long VAL,int Q = 0) { ma = MA; mb = MB; mc = MC; n = NN; q = Q; val = VAL; } bool operator==(const node & B) { return (ma == B.ma && mb == B.mb && mc == B.mc && q == B.q && n == B.n); }}; std::queue<node> queue; Long f[N + 10][N + 10][N + 10][N + 10][2]; Long res = INF; int main(){ scanf("%lld%lld%lld",&a,&b,&c); na = get_num(a); nb = get_num(b); nc = get_num(c); int lena = get_len(a); int lenb = get_len(b); int lenc = get_len(c); len = lena; if(len < lenb) len = lenb; if(len < lenc) len = lenc; memset(f,-1,sizeof(f)); queue.push(node(0,0,0,0,0,0)); f[0][0][0][0][0] = 0; #define F(A) (f[A.ma][A.mb][A.mc][A.n][A.q]) #define UP() \ {\ if(now_s.val < F(now_s) || F(now_s) < 0)\ {\ F(now_s) = now_s.val;\ queue.push(now_s);\ }\ } node now_f; node now_s; do { now_f = queue.front(); queue.pop(); if(F(now_f) < now_f.val) continue; //if(now_f == node(2,2,2 ,2 , 1,1)) // fprintf(stderr,"QAQ %d\n",now_f.val); // if(now_f.n == 2) // fprintf(stderr,"%d %d %d n = %d val = %lld\n",now_f.ma,now_f.mb,now_f.mc,now_f.n,now_f.val); if(now_f.n > len) continue; { now_s = now_f; now_s.n++; now_s.q = 0; UP(); } if(now_f.ma < na) { now_s = now_f; now_s.ma++; now_s.n++; if(now_s.q == 1) { now_s.val -= pow(2,now_s.n-1); now_s.val += pow(2,now_s.n); } else { now_s.val += pow(2,now_s.n-1); now_s.mc++; } UP(); } if(now_f.mb < nb) { now_s = now_f; now_s.mb++; now_s.n++; if(now_s.q == 1) { now_s.val -= pow(2,now_s.n-1); now_s.val += pow(2,now_s.n); } else { now_s.val += pow(2,now_s.n-1); now_s.mc++; } UP(); } if(now_f.ma < na && now_f.mb < nb) { now_s = now_f; now_s.mb++; now_s.ma++; now_s.n++; if(now_s.q == 1) { now_s.val += pow(2,now_s.n); now_s.mc ++; } else { now_s.val += pow(2,now_s.n); now_s.mc++; now_s.q = 1; } UP(); } } while(!queue.empty()); upmin(res,f[na][nb][nc][len][0]);// upmin(res,f[na][nb][nc][len][1]); // fprintf(stderr,"%lld %lld\n",f[na][nb][nc][len][0],f[na][nb][nc][len][1]); if(res == INF) res = -1LL; printf("%lld\n",res); return 0;} |
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