# 题目

{% fold 点击显/隐题目 %}

1 100 0 0
80
{% endfold %}

# 题解

`dp[i][j] = sum{ dp[i-1][k] }` 其中 `k` 是满足条件的数

# 代码

{% fold 点击显/隐代码 %}```cpp 不要62 https://github.com/OhYee/sourcecode/tree/master/ACM 代码备份
#include
#include
#include
using namespace std;

const int maxn = 10;

int a[maxn];
int dp[maxn][10];

/* 第pos位 前一位状态为 pre 当前是否限制当前位(limit) */
int dfs(int pos, int pre, bool limit) {
//for (int i = maxn; i >= pos; i--)
// printf("\t");
//printf("dfs(%d,%d,%d,%d)\n", pos, pre, limit);

``````if (pos == -1)
return 1;

if (limit || dp[pos][pre] == -1) {
int up = limit ? a[pos] : 9;
int ans = 0;
for (int now = 0; now <= up; now++) {
if (pre == 6 && now == 2)
continue;
if (now == 4)
continue;
ans += dfs(pos - 1, now, limit && now == a[pos]);
}
if (!limit)
dp[pos][pre] = ans;
return ans;
}
return dp[pos][pre];
``````

}

int solve(int x) {
int pos = 0;
while (x) {
a[pos++] = x % 10;
x /= 10;
}
int ans = dfs(pos - 1, -1, true);
//printf("solve(%d) = %d\n", x, ans);
return ans;
}
int main() {
int l, r;
while (scanf("%d%d", &l, &r) != EOF) {
if (l == 0 && r == 0)
break;
memset(dp, -1, sizeof(dp));
printf("%d\n", solve(r) - solve(l - 1));
}
return 0;
}

`{% endfold %}`