HDU 4249 A Famous Equation(数位DP)

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题目链接:点击打开链接

思路:用d[i][a][b][c][is]表示当前到了第i位, 三个数的i位各自是a,b,c, 是否有进位 , 的方法数。

细节參见代码:

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<string>
#include<vector>
#include<stack>
#include<bitset>
#include<cstdlib>
#include<cmath>
#include<set>
#include<list>
#include<deque>
#include<map>
#include<queue>
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
using namespace std;
typedef long long ll;
typedef long double ld;
const ld eps = 1e-9, PI = 3.1415926535897932384626433832795;
const int mod = 1000000000 + 7;
const int INF = 0x3f3f3f3f;
// & 0x7FFFFFFF
const int seed = 131;
const ll INF64 = ll(1e18);
const int maxn = 15;
int T,n,m,len,vis[maxn][maxn][maxn][maxn][2],len1,len2,len3,kase = 0;
char a[maxn],b[maxn],c[maxn],s[100];
ll d[maxn][maxn][maxn][maxn][2];
ll dp(int pos, int bb, int cc, int dd, int is) {
    ll& ans = d[pos][bb][cc][dd][is];
    if(pos > len) return is == 0;
    if(vis[pos][bb][cc][dd][is] == kase) return ans;
    vis[pos][bb][cc][dd][is] = kase;
    ans = 0;
        if(a[pos] == ‘?‘ && b[pos] == ‘?‘) {
            for(int i = 0; i < 10; i++) {
                for(int j = 0; j < 10; j++) {
                    if(pos == len1 && i == 0 && len1 != 1) continue;
                    if(pos == len2 && j == 0&& len2 != 1) continue;
                    int cur = i + j + is;
                    int res = 0;
                    if(cur >= 10) {
                        cur -= 10; ++res;
                    }
                    if(c[pos] == ‘?

‘ && !(pos == len3 && cur == 0 && len3 != 1)) ans += dp(pos+1,i, j,cur, res); else if(c[pos]-‘0‘ == cur) ans += dp(pos+1, i,j,cur, res); } } } else if(a[pos] == ‘?‘) { for(int i = 0; i < 10; i++) { if(pos == len1 && i == 0&& len1 != 1) continue; int cur = i + b[pos]-‘0‘ + is; int res = 0; if(cur >= 10) { cur -= 10; ++res; } if(c[pos] == ‘?

‘ && !(pos == len3 && cur == 0 && len3 != 1)) ans += dp(pos+1,i,b[pos]-‘0‘,cur, res); else if(c[pos]-‘0‘ == cur) ans += dp(pos+1,i,b[pos]-‘0‘,cur, res); } } else if(b[pos] == ‘?‘) { for(int i = 0; i < 10; i++) { if(pos == len2 && i == 0&& len2 != 1) continue; int cur = i + a[pos]-‘0‘ + is; int res = 0; if(cur >= 10) { cur -= 10; ++res; } if(c[pos] == ‘?‘ && !(pos == len3 && cur == 0 && len3 != 1)) ans += dp(pos+1,a[pos]-‘0‘,i,cur, res); else if(c[pos]-‘0‘ == cur) ans += dp(pos+1,a[pos]-‘0‘,i,cur, res); } } else { int cur = a[pos]-‘0‘ + b[pos]-‘0‘ + is; int res = 0; if(cur >= 10) { cur -= 10; ++res; } if(c[pos] == ‘?‘ && !(pos == len3 && cur == 0 && len3 != 1)) ans += dp(pos+1, a[pos]-‘0‘,b[pos]-‘0‘,cur, res); else if(c[pos]-‘0‘ == cur) ans += dp(pos+1, a[pos]-‘0‘,b[pos]-‘0‘,cur, res); } return ans; } int main() { while(~scanf("%s",s+1)) { len = strlen(s+1); len1 = 0; len2 = 0; len3 = 0; int id = 0; for(int i = len; i >= 1; i--) { if(s[i] == ‘=‘ || s[i] == ‘+‘) { id++; continue; } if(id == 0) { c[++len3] = s[i]; } else if(id == 1) { b[++len2] = s[i]; } else { a[++len1] = s[i]; } } len = max(len1, max(len2, len3));//补全不足的。 降低代码量 for(int i = len1+1; i <= len; i++) a[i] = ‘0‘; for(int i = len2+1; i <= len; i++) b[i] = ‘0‘; for(int i = len3+1; i <= len; i++) c[i] = ‘0‘; ++kase; ll ans = dp(1, 0 , 0, 0, 0); printf("Case %d: %I64d\n",kase,ans); } return 0; }



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