一眼數位dp，但。。。調試了好久。。

/*** Problem:Count* Author:Shun Yao* Time:2013.5.31* Result:Accepted* Memo:DP*/#include <cstdio> #include <cstring>long long ans[10], s[15], tmp[10], f[15][10][10];void calc(long long b, char y) {long i, j, l, t, k;long long q;if (b == 0) {if (y)++ans[0];else--ans[0];return;}memset(s, 0, sizeof s);memset(f, 0, sizeof f);memset(tmp, 0, sizeof tmp);k = 0;while (b) {s[++k] = b % 10;b /= 10;}q = 1;for (i = 1; i <= k; ++i) {for (j = 0; j <= 9; ++j) {f[i][j][j] = q;for (l = 0; l <= 9; ++l)for (t = 0; t <= 9; ++t)f[i][j][l] += f[i - 1][t][l];}q *= 10;}if (k > 0) {for (j = 2; j <= k; ++j)tmp[s[j]] += s[1] + 1;for (l = 0; l <= 9; ++l)for (j = 0; j <= s[1]; ++j)tmp[l] += f[1][j][l];}q = 10;for (i = 2; i < k; ++i) {for (j = i + 1; j <= k; ++j)tmp[s[j]] += s[i] * q;for (l = 0; l <= 9; ++l)for (j = 0; j < s[i]; ++j)tmp[l] += f[i][j][l];q *= 10;}if (k > 2) {for (i = 0; i <= 9; ++i)tmp[i] += f[1][i][i];for (i = 2; i < k; ++i)for (j = 0; j <= 9; ++j)for (l = 1; l <= 9; ++l)tmp[j] += f[i][l][j];for (i = 1; i < s[k]; ++i)for (j = 0; j <= 9; ++j)tmp[j] += f[k][i][j];}for (i = 0; i <= 9; ++i)if (y)ans[i] = tmp[i];elseans[i] -= tmp[i]; }int main() {long i;long long a, b;freopen("count.in", "r", stdin);freopen("count.out", "w", stdout);scanf("%lld%lld", &a, &b);calc(b, 1);calc(a - 1, 0);printf("%lld", ans[0]);for (i = 1; i <= 9; ++i)printf(" %lld", ans[i]);fclose(stdin);fclose(stdout);return 0; }

?

轉載于:https://www.cnblogs.com/hsuppr/archive/2013/06/03/3115477.html

總結

以上是生活随笔為你收集整理的[ZJOI2010]数字计数的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。