【HDOJ 3555】Bomb

【HDOJ 3555】Bomb

基础数位dp 跟那个4 62差不多 这个能更简单点 慢慢来吧= 。= 预处理好49连续的情况即可

ps:补一发记忆化的。。用推的15ms 记忆化109…有毒TOT

代码如下:

#include <iostream>#include <cstdio>#include <cstring>#define ll long long#define sc "%I64d"using namespace std;ll dp[20][3];/*dp[0] 无连续49dp[1] 无连续49高位为9dp[2] 有连续49*/void Init(){    memset(dp,0,sizeof(dp));    dp[0][0] = 1;    int i;    for(i = 1; i <= 19; ++i)    {        dp[i][0] = dp[i-1][0]*10 - dp[i-1][1];        dp[i][1] = dp[i-1][0];        dp[i][2] = dp[i-1][2]*10 + dp[i-1][1];    }}ll Solve(ll n){    int len = 0,bit[21],i,f = false;    ll ans = 0;    while(n)    {        bit[++len] = n%10;        n /= 10;    }    bit[len+1] = 0;    for(i = len; i; --i)    {        ans += dp[i-1][2]*bit[i];        if(f) ans += dp[i-1][0]*bit[i];        if(!f && bit[i] > 4) ans += dp[i-1][1];        if(bit[i+1] == 4 && bit[i] == 9) f = true;    }    return ans;}int main(){    int t;    ll n;    scanf("%d",&t);    Init();    while(t--)    {        scanf(sc,&n);        printf(sc,Solve(n+1));        puts("");    }    return 0;}

#include <iostream>#include <cstdio>#include <cstring>#define ll long longusing namespace std;ll dp[20][3];int digit[20];/*hav =0 没有连续491 没有连续49前位为42 有连续49high 高位是否出现 未出现则随意填*/ll dfs(int pos,int hav,bool high){    if(pos == -1) return hav == 2;    if(!high && ~dp[pos][hav]) return dp[pos][hav];    int i,nhav,en;    ll ans = 0;    en = high? digit[pos]: 9;    for(i = 0; i <= en; ++i)    {        nhav = hav;        if(!nhav && i == 4) nhav = 1;        else if(nhav == 1 && i == 9) nhav = 2;        else if(nhav == 1 && i != 4) nhav = 0;        ans += dfs(pos-1,nhav,high && i == en);    }    if(!high) dp[pos][hav] = ans;    return ans;}ll Solve(char *x){    memset(dp,-1,sizeof(dp));    int len = strlen(x);    int i;    for(i = 0; i < len; ++i)        digit[i] = x[len-i-1]-'0';    return dfs(len-1,0,1);}int main(){    int t;    char n[21];    scanf("%d",&t);    while(t--)    {        scanf("%s",n);        printf("%I64d\n",Solve(n));    }    return 0;}