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题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5898
题意:求[l,r]区间内数字,满足连续奇数的个数是偶数个,连续偶数的个数是奇数个。
dp(l,pre,con,fz)表示前l位,最后一位是pre,并且此时这个pre所在的连通块已经有con个了,fz来区分是不是前导零。
1 #include <bits/stdc++.h> 2 using namespace std; 3 #define fr first 4 #define sc second 5 #define cl clear 6 #define BUG puts("here!!!") 7 #define W(a) while(a--) 8 #define pb(a) push_back(a) 9 #define Rint(a) scanf("%d", &a) 10 #define Rll(a) scanf("%I64d", &a) 11 #define Rs(a) scanf("%s", a) 12 #define Cin(a) cin >> a 13 #define FRead() freopen("in", "r", stdin) 14 #define FWrite() freopen("out", "w", stdout) 15 #define Rep(i, len) for(int i = 0; i < (len); i++) 16 #define For(i, a, len) for(int i = (a); i < (len); i++) 17 #define Cls(a) memset((a), 0, sizeof(a)) 18 #define Clr(a, x) memset((a), (x), sizeof(a)) 19 #define Full(a) memset((a), 0x7f7f7f, sizeof(a)) 20 #define lrt rt << 1 21 #define rrt rt << 1 | 1 22 #define pi 3.14159265359 23 #define RT return 24 #define lowbit(x) x & (-x) 25 #define onecnt(x) __builtin_popcount(x) 26 typedef long long LL; 27 typedef long double LD; 28 typedef unsigned long long ULL; 29 typedef pair<int, int> pii; 30 typedef pair<string, int> psi; 31 typedef pair<LL, LL> pll; 32 typedef map<string, int> msi; 33 typedef vector<int> vi; 34 typedef vector<LL> vl; 35 typedef vector<vl> vvl; 36 typedef vector<bool> vb; 37 38 const int maxn = 19; 39 int digit[maxn]; 40 LL l, r; 41 LL dp[maxn][2][2][2]; 42 43 LL dfs(int l, int pre, int con, bool fz, bool flag) { 44 if(l == 0) { 45 if(pre % 2 == 0) { 46 if(con % 2 == 0) return 0; 47 if(con % 2 == 1) return 1; 48 } 49 else { 50 if(con % 2 == 0) return 1; 51 if(con % 2 == 1) return 0; 52 } 53 } 54 if(!flag && ~dp[l][pre][con][fz]) return dp[l][pre][con][fz]; 55 LL ret = 0; 56 int pos = flag ? digit[l] : 9; 57 if(fz) { 58 Rep(i, pos+1) { 59 ret += dfs(l-1, i%2, 1, fz&&(i==0), flag&&(i==pos)); 60 } 61 } 62 else if((pre + con) % 2 == 1) { 63 Rep(i, pos+1) { 64 if((i + pre) % 2 == 0) { 65 ret += dfs(l-1, i%2, (con+1)%2, fz&&(i==0), flag&&(i==pos)); 66 } 67 else { 68 ret += dfs(l-1, i%2, 1, fz&&(i==0), flag&&(i==pos)); 69 } 70 } 71 } 72 else { 73 Rep(i, pos+1) { 74 if((i + pre) % 2 == 0) { 75 ret += dfs(l-1, i%2, (con+1)%2, fz&&(i==0), flag&&(i==pos)); 76 } 77 } 78 } 79 if(!flag) dp[l][pre][con][fz] = ret; 80 return ret; 81 } 82 83 LL f(LL x) { 84 int pos = 0; 85 while(x) { 86 digit[++pos] = x % 10; 87 x /= 10; 88 } 89 return dfs(pos, 0, 1, true, true); 90 } 91 92 signed main() { 93 //FRead(); 94 int T, _ = 1; 95 Rint(T); 96 Clr(dp, -1); 97 W(T) { 98 cin >> l >> r; 99 printf("Case #%d: ", _++); 100 cout << f(r) - f(l-1) << endl; 101 } 102 RT 0; 103 }
[HDOJ5898]odd-even number(数位dp)
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原文地址:http://www.cnblogs.com/vincentX/p/5895624.html