HDU 5898 odd-even number(2016 ACM/ICPC Asia Regional Shenyang Online)

题目分析

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;

LL dp[20][20][20], bit[20];

LL dfs(int pos, bool zero, int ji, int ou, int limit){
if(pos < 1){ //如果到末位
if(!ji&&!ou) return 1;
else if(ji && !ou) return ji%2==0;
else if(!ji && ou) return ou&1;
else return (ji%2==0)&&(ou&1);
}
if(!limit && dp[pos][ji][ou] != -1) return dp[pos][ji][ou];
LL ret = 0;
int len = limit?bit[pos]:9;
for(int i = 0; i <= len; i++){
if(i&1){  //如果为奇数
if((ou&1)||(ou==0)) ret += dfs(pos-1, false, ji+1, 0, limit&&i==len);
}
else{
if(i == 0){ //如果该位置为0
if(zero){ //且有前导0
ret += dfs(pos-1, true, 0, 0, limit&&i==len);
}
else if(ji%2==0){
ret += dfs(pos-1, false, 0, ou+1, limit&&i==len);
}
}
else{
if(ji%2==0) ret += dfs(pos-1, false, 0, ou+1, limit&&i==len);
}
}
}
if(!limit) dp[pos][ji][ou] = ret;
return ret;
}

LL solve(LL x){
int len = 0;
while(x) bit[++len] = x%10, x /= 10;
return dfs(len, true, 0, 0, true);
}

int main(){
int T;
scanf("%d", &T);
LL L, R;
memset(dp, -1, sizeof(dp));
for(int kase = 1; kase <= T; kase++){
scanf("%I64d%I64d", &L, &R);
printf("Case #%d: %I64d\n", kase, solve(R)-solve(L-1));
}
return 0;
}