题目链接：
http://acm.hdu.edu.cn/showproblem.php?pid=5898
odd-even number
Memory Limit: 65536/65536 K (Java/Others)
输入
First line a t,then t cases.every line contains two integers L and R.
输出
Print the output for each case on one line in the format as shown below.
样例输入
2
1 100
110 220
样例输出
Case #1: 29
Case #2: 36
题意
求[l,r]区间里面满足任意连续个数位为偶数的长度为奇数，任意连续个数位为奇数的长度为偶数，比如110,1100033334等。
题解
数位dp具体看代码注释
#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<ctime>
#include<vector>
#include<cstdio>
#include<string>
#include<bitset>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
#define mid (l+(r-l)/2)
#define sz() size()
#define pb(v) push_back(v)
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)
#define scf scanf
#define prf printf
typedef long long LL;
typedef vector<int> VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;
const int INF=0x3f3f3f3f;
const LL INFL=0x3f3f3f3f3f3f3f3fLL;
const double eps=1e-6;
const double PI = acos(-1.0);
//start----------------------------------------------------------------------
const int maxn=22;
int arr[maxn],tot;
LL dp[maxn][2][2];
///dp[i][0][0]表示长度为i，最后偶数位为偶数的情况
///dp[i][0][1]表示长度为i，最后奇数位为偶数的情况
///dp[i][1][0]表示长度为i，最后偶数位为奇数的情况
///dp[i][1][1]表示长度为i，最后奇数位为奇数的情况
LL dfs(int len,int x,int y,bool ismax,bool iszer) {
if (len == 0) {
///递归边界
return y^1;
}
if (!ismax&&!iszer&&dp[len][x][y]>=0) return dp[len][x][y];
LL res = 0;
int ed = ismax ? arr[len] : 9;
///这里写递推的代码
if(x==0){
for(int i=0;i<=ed;i+=2){
if(i==0&&iszer){
///处理前导零
res+=dfs(len-1,0,1,ismax&&i==ed,iszer&&i==0);
res+=dfs(len-1,1,0,ismax&&i==ed,iszer&&i==0);
}
else{
///后面接的还是偶数的情况
res+=dfs(len-1,0,y^1,ismax&&i==ed,iszer&&i==0);
///后面能接奇数的情况
if(y==1&&len>1) res+=dfs(len-1,1,0,ismax&&i==ed,iszer&&i==0);
}
}
}else{
for(int i=1;i<=ed;i+=2){
///后面接的还是奇数的情况
res+=dfs(len-1,1,y^1,ismax&&i==ed,iszer&&i==0);
///后面接的是偶数的情况
if(y==1&&len>1) res+=dfs(len-1,0,1,ismax&&i==ed,iszer&&i==0);
}
}
///这里记忆化存的必须保证没有前导零的情况！，有前导零的要单独计算
return ismax||iszer ? res : dp[len][x][y] = res;
}
LL solve(LL x) {
tot = 0;
while (x) { arr[++tot] = x % 10; x /= 10; }
LL even=dfs(tot, 0,1, true,true);
LL odd=dfs(tot,1,0,true,true);
return even+odd;
}
void init(){
clr(dp,-1);
}
int main() {
int tc,kase=0;
scf("%d",&tc);
init();
while(tc--){
LL x,y;
scf("%lld%lld",&x,&y);
prf("Case #%d: %lld
",++kase,solve(y)-solve(x-1));
}
return 0;
}
//end-----------------------------------------------------------------------
再回顾一发：
#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<ctime>
#include<vector>
#include<cstdio>
#include<string>
#include<bitset>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
#define mid (l+(r-l)/2)
#define sz() size()
#define pb(v) push_back(v)
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)
#define scf scanf
#define prf printf
typedef long long LL;
typedef vector<int> VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;
const int INF=0x3f3f3f3f;
const LL INFL=10000000000000000LL;
const double eps=1e-9;
const double PI = acos(-1.0);
//start----------------------------------------------------------------------
LL dp[22][3][2];
int arr[22],tot;
///k==0表示上一位不存在，k==1表示上一位为奇数，k==2表示上一位为偶数
///l==0表示上一位连续偶数个奇数或偶数的情况
LL dfs(int len,int k,int l, bool ismax,bool iszer) {
if (len == 0) {
if(l==0) return 1LL;
return 0;
}
if (!ismax&&dp[len][k][l]>=0) return dp[len][k][l];
LL res = 0;
int ed = ismax ? arr[len] : 9;
///这里插入递推公式
for (int i = 0; i <= ed; i++) {
if(i==0&&iszer){
///处理前导零
res+=dfs(len-1,0,0,ismax&&i==ed,iszer&&i==0);
}else{
if(k==0){
if(i&1) res+=dfs(len-1,1,1,ismax&&i==ed,iszer&&i==0);
else res+=dfs(len-1,2,0,ismax&&i==ed,iszer&&i==0);
}else if(k==1){
if(i&1){
res+=dfs(len-1,k,l^1,ismax&&i==ed,iszer&&i==0);
}else{
if(l==0) res+=dfs(len-1,2,0,ismax&&i==ed,iszer&&i==0);
}
}else if(k==2){
if(!(i&1)){
res+=dfs(len-1,k,l^1,ismax&&i==ed,iszer&&i==0);
}else{
if(l==0) res+=dfs(len-1,1,1,ismax&&i==ed,iszer&&i==0);
}
}
}
}
return ismax ? res : dp[len][k][l] = res;
}
LL solve(LL x) {
tot = 0;
while (x) { arr[++tot] = x % 10; x /= 10; }
return dfs(tot,0,0,true,true);
}
int main() {
clr(dp,-1);
int tc,kase=0;
scf("%d",&tc);
while(tc--){
LL l,r;
scf("%lld%lld",&l,&r);
prf("Case #%d: %lld
",++kase,solve(r)-solve(l-1));
}
return 0;
}
//end-----------------------------------------------------------------------
精简版：
#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<ctime>
#include<vector>
#include<cstdio>
#include<string>
#include<bitset>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
#define mid (l+(r-l)/2)
#define sz() size()
#define pb(v) push_back(v)
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)
#define scf scanf
#define prf printf
typedef long long LL;
typedef vector<int> VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;
const int INF=0x3f3f3f3f;
const LL INFL=10000000000000000LL;
const double eps=1e-9;
const double PI = acos(-1.0);
//start----------------------------------------------------------------------
LL dp[22][3][2];
int arr[22],tot;
///k==0表示上一位不存在，k==1表示上一位为奇数，k==2表示上一位为偶数
///l==0表示上一位连续偶数个奇数或偶数的情况
LL dfs(int len,int k,int l, bool ismax,bool iszer) {
if (len == 0) return l^1;
if (!ismax&&dp[len][k][l]>=0) return dp[len][k][l];
LL res = 0;
int ed = ismax ? arr[len] : 9;
for (int i = 0; i <= ed; i++) {
if(i==0&&iszer){
res+=dfs(len-1,2,0,ismax&&i==ed,iszer&&i==0);
}else{
if(k==2){
if(i&1) res+=dfs(len-1,1,1,ismax&&i==ed,iszer&&i==0);
else res+=dfs(len-1,0,0,ismax&&i==ed,iszer&&i==0);
}else{
if(!(k^(i&1))) res+=dfs(len-1,k,l^1,ismax&&i==ed,iszer&&i==0);
else if(!l) res+=dfs(len-1,!k,!k,ismax&&i==ed,iszer&&i==0);
}
}
}
return ismax ? res : dp[len][k][l] = res;
}
LL solve(LL x) {
tot = 0;
while (x) { arr[++tot] = x % 10; x /= 10; }
return dfs(tot,2,0,true,true);
}
int main() {
clr(dp,-1);
int tc,kase=0;
scf("%d",&tc);
while(tc--){
LL l,r;
scf("%lld%lld",&l,&r);
prf("Case #%d: %lld
",++kase,solve(r)-solve(l-1));
}
return 0;
}
//end-----------------------------------------------------------------------
Notes
数位dp，比赛的时候思路歪了，干了两个多小时没出样例。。第二天条整思路之后，又调了一个上午。。。。orz，dp弱渣。。。