下面再放三道我比較喜歡的，需要好好寫一下的題。

第一題比較水

1. White Cloud is exercising in the playground. White Cloud can walk 1 meters or run k meters per second. Since White Cloud is tired,it can't run for two or more continuous seconds. White Cloud will move L to R meters. It wants to know how many different ways there are to achieve its goal. Two ways are different if and only if they move different meters or spend different seconds or in one second, one of them walks and the other runs.

輸入描述： The first line of input contains 2 integers Q and k.Q is the number of queries.(Q<=100000,1<=k<=100000) For the next Q lines,each line contains two integers L and R.(1<=L<=R<=100000)

輸出描述： For each query,print a line which contains an integer,denoting the answer of the query modulo 1000000007.

示例1: 輸入 3 3 3 3 1 4 1 5 輸出 2 7 11

題目大意

白云在健身，每秒可以走1米或跑k米，并且不能連續(xù)兩秒都在跑。 當(dāng)它的移動(dòng)距離在[L,R]之間時(shí)，可以選擇結(jié)束鍛煉。 問有多少種方案結(jié)束。

做法

DP? f[i][0/1]表示已經(jīng)過了i米，最后一步是跑還是走的方案數(shù)，分開就很好寫了，簡(jiǎn)單，自己體會(huì)。 f[i][1]=f[i-k][0],f[i][0]=f[i-1][0]+f[i-1][1] 。注意要記錄前綴和。

?

#include <bits/stdc++.h> #define MOD 1000000007 using namespace std; typedef long long ll; ll g[100005]; ll dp[100005][2]; int main(void) {int q,k,a,b;ll sum;cin>>q>>k;dp[0][0]=1;for(int i=1;i<=100000;++i){dp[i][0]=(dp[i-1][0]+dp[i-1][1])%MOD;if(i-k>=0)dp[i][1]=(dp[i-k][0])%MOD;}sum=0;for(int i=1;i<=100000;++i){g[i]=(dp[i][0]+dp[i][1]+sum);sum=g[i];}// for(int i=9000;i<=100000;++i){// cout<<g[i]<<endl;// }while(q--){cin>>a>>b;cout<<(g[b]-g[a-1])%MOD<<endl;}return 0; }

?

2、來一發(fā)多重背包

選這個(gè)題是感覺代碼有一種美感。。。。。

一. 編程題

1. Since then on, Eddy found that physics is actually the most important thing in the contest. Thus, he wants to form a team to guide the following contestants to conquer the PACM contests(PACM is short for Physics, Algorithm, Coding, Math).

There are N candidate groups each composed of pi physics experts, ai algorithm experts, ci coding experts, mi math experts. For each group, Eddy can either invite all of them or none of them. If i-th team is invited, they will bring gi knowledge points which is calculated by Eddy's magic formula. Eddy believes that the higher the total knowledge points is, the better a team could place in a contest. But, Eddy doesn't want too many experts in the same area in the invited groups. Thus, the number of invited physics experts should not exceed P, and A for algorithm experts, C for coding experts, M for math experts.

Eddy is still busy in studying Physics. You come to help him to figure out which groups should be invited such that they doesn't exceed the constraint and will bring the most knowledge points in total.

輸入描述： The first line contains a positive integer N indicating the number of candidate groups. Each of following N lines contains five space-separated integer p i, ai, ci, mi, gi indicating that i-th team consists of pi physics experts, ai algorithm experts, ci coding experts, mi math experts, and will bring gi knowledge points. The last line contains four space-separated integer P, A, C, M indicating the maximum possible number of physics experts, algorithm experts, coding experts, and math experts, respectively.

?1 ≤ N ≤ 36 0 ≤ pi,ai,ci,mi,gi ≤ 36 0 ≤ P, A, C, M ≤ 36

輸出描述： The first line should contain a non-negative integer K indicating the number of invited groups. The second line should contain K space-separated integer indicating the index of invited groups(groups are indexed from 0).

You can output index in any order as long as each index appears at most once. If there are multiple way to reach the most total knowledge points, you can output any one of them. If none of the groups will be invited, you could either output one line or output a blank line in the second line.

題干有點(diǎn)長(zhǎng)，其實(shí)就是每個(gè)隊(duì)有四種代價(jià)，一個(gè)價(jià)值。多重背包。36的五次方竟然沒超時(shí)也是醉了。。

?

#include <bits/stdc++.h> using namespace std; const int N=37; int n, p[N], a[N], c[N], m[N], g[N]; int P, A, C, M; short dp[N][N][N][N][N]; bool tk[N][N][N][N][N]; int main(){cin>>n;for(int i=0; i<n; i++)cin>>p[i]>>a[i]>>c[i]>>m[i]>>g[i];cin>>P>>A>>C>>M;for(int i=0; i<=n; i++)for(int ip=0; ip<=P; ip++)for(int ia=0; ia<=A; ia++)for(int ic=0; ic<=C; ic++)for(int im=0; im<=M; im++)tk[i][ip][ia][ic][im]=false;for(int i=0; i<n; i++){for(int ip=P; ip>=0; ip--)for(int ia=A; ia>=0; ia--)for(int ic=C; ic>=0; ic--)for(int im=M; im>=0; im--)dp[i+1][ip][ia][ic][im]=dp[i][ip][ia][ic][im];for(int ip=P; ip>=p[i]; ip--)for(int ia=A; ia>=a[i]; ia--)for(int ic=C; ic>=c[i]; ic--)for(int im=M; im>=m[i]; im--)if(dp[i][ip-p[i]][ia-a[i]][ic-c[i]][im-m[i]]+g[i]>dp[i+1][ip][ia][ic][im]){dp[i+1][ip][ia][ic][im]=dp[i][ip-p[i]][ia-a[i]][ic-c[i]][im-m[i]]+g[i];tk[i+1][ip][ia][ic][im]=true;}}fprintf(stderr, "ans=%d\n", dp[n][P][A][C][M]);vector<int> ans;for(int i=n; i>=1; i--)if(tk[i][P][A][C][M]){ans.push_back(i-1);P-=p[i-1];A-=a[i-1];C-=c[i-1];M-=m[i-1];}reverse(ans.begin(), ans.end());printf("%d\n", (int)ans.size());for(size_t i=0; i<ans.size(); i++)printf("%d%c", ans[i], " \n"[i+1==ans.size()]); }

3、P1040 加分二叉樹

題目

https://www.luogu.org/problemnew/show/P1040

設(shè)一個(gè)?nn?個(gè)節(jié)點(diǎn)的二叉樹tree的中序遍歷為（?1,2,3,…,n1,2,3,…,n?），其中數(shù)字?1,2,3,…,n1,2,3,…,n?為節(jié)點(diǎn)編號(hào)。每個(gè)節(jié)點(diǎn)都有一個(gè)分?jǐn)?shù)（均為正整數(shù)），記第?ii?個(gè)節(jié)點(diǎn)的分?jǐn)?shù)為?di,treedi,tree?及它的每個(gè)子樹都有一個(gè)加分，任一棵子樹?subtreesubtree?（也包含?treetree?本身）的加分計(jì)算方法如下：

subtreesubtree?的左子樹的加分×?subtreesubtree?的右子樹的加分＋?subtreesubtree?的根的分?jǐn)?shù)。

若某個(gè)子樹為空，規(guī)定其加分為?11?，葉子的加分就是葉節(jié)點(diǎn)本身的分?jǐn)?shù)。不考慮它的空子樹。

試求一棵符合中序遍歷為（?1,2,3,…,n1,2,3,…,n?）且加分最高的二叉樹?treetree?。要求輸出；

（1）?treetree?的最高加分

（2）?treetree?的前序遍歷

思路：這個(gè)題可以用動(dòng)態(tài)規(guī)劃或者記憶化搜索來做。因?yàn)槿绻蠹臃肿畲蟮脑?#xff0c;必須要求它的兒子結(jié)點(diǎn)加分最大，所以就有了最優(yōu)子階段。我們可以枚舉根來更新最大值。

root[i,j]表示[i,j]這段序列的根，遞歸輸出先序遍歷。注意初始化，f[i][i]=v[i]，當(dāng)序列只有I一個(gè)元素時(shí)，f[i][i]等于這個(gè)點(diǎn)本身的權(quán)值，當(dāng)l==r-1時(shí)，此時(shí)是空樹設(shè)為1。

?

#include<iostream> #include<cstdio> using namespace std; int n,v[39],f[47][47],i,j,k,root[49][49]; void print(int l,int r){if(l>r)return;if(l==r){printf("%d ",l);return;}printf("%d ",root[l][r]);print(l,root[l][r]-1);print(root[l][r]+1,r); } int main() {scanf("%d",&n);for( i=1; i<=n; i++) scanf("%d",&v[i]);for(i=1; i<=n; i++) {f[i][i]=v[i];f[i][i-1]=1;}for(i=n; i>=1; i--)for(j=i+1; j<=n; j++)for(k=i; k<=j; k++) {if(f[i][j]<(f[i][k-1]*f[k+1][j]+f[k][k])) {f[i][j]=f[i][k-1]*f[k+1][j]+f[k][k];root[i][j]=k;}}printf("%d\n",f[1][n]);print(1,n);return 0; }

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