HDU 4722 Good Numbers
2013-09-11 17:22
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题意:求出给定范围内的“Good Numbers”,也就是所有数位之和可以整除10即可
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4722
思路:数位dp
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
using namespace std;
#define LL long long
LL digit[30];
LL dp[30][30];
LL dfs(LL pos , LL pre , bool doing){
if(pos == -1) return pre == 0;
if(!doing && dp[pos][pre]!=-1)
return dp[pos][pre];
LL ans = 0 , npre;
LL end = doing ? digit[pos] : 9;
for(LL i = 0 ; i <= end ; i ++){
npre = (pre + i) % 10;
ans += dfs(pos - 1, npre , doing && (i == end));
}
if(!doing) dp[pos][pre] = ans;
return ans;
}
LL calc(LL x){
LL pos = 0;
memset(dp , -1,sizeof(dp));
while(x){
digit[pos ++]= x % 10;
x /= 10;
}
return dfs(pos - 1, 0 , 1);
}
LL a,b,ans;
int main()
{
int T;
scanf("%d",&T);
for(int ncase = 1;ncase <= T;ncase++){
scanf("%I64d %I64d",&a,&b);
ans = calc(b) - calc(a - 1);
printf("Case #%d: %I64d\n",ncase,ans);
}
return 0;
}
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4722
思路:数位dp
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
using namespace std;
#define LL long long
LL digit[30];
LL dp[30][30];
LL dfs(LL pos , LL pre , bool doing){
if(pos == -1) return pre == 0;
if(!doing && dp[pos][pre]!=-1)
return dp[pos][pre];
LL ans = 0 , npre;
LL end = doing ? digit[pos] : 9;
for(LL i = 0 ; i <= end ; i ++){
npre = (pre + i) % 10;
ans += dfs(pos - 1, npre , doing && (i == end));
}
if(!doing) dp[pos][pre] = ans;
return ans;
}
LL calc(LL x){
LL pos = 0;
memset(dp , -1,sizeof(dp));
while(x){
digit[pos ++]= x % 10;
x /= 10;
}
return dfs(pos - 1, 0 , 1);
}
LL a,b,ans;
int main()
{
int T;
scanf("%d",&T);
for(int ncase = 1;ncase <= T;ncase++){
scanf("%I64d %I64d",&a,&b);
ans = calc(b) - calc(a - 1);
printf("Case #%d: %I64d\n",ncase,ans);
}
return 0;
}
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