哈尔滨理工大学软件学院ACM程序设计全国邀请赛（网络同步赛）

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C

//题目给出n，表示n个硬币，能够凑成所有0～m的数，并且这n个硬币的和是m//那么定义撞塌dp[i][j][k]，i代表硬币数，j代表能表示1～j的数，k表示硬币中最大的数#include<iostream>using namespace std;typedef long long LL;const LL mod=2*1e9+7;const int maxn=200+5;LL f[maxn][maxn][maxn];int main(){f[1][1][1]=1;for(int i=1;i<=200;i++){for(int j=1;j<=200;j++){for(int k=1;k<=200;k++){if(!f[i][j][k]) continue;for(int add=k;add+j<=200&&add<j+2;add++){f[i+1][j+add][add]+=f[i][j][k];f[i+1][j+add][add]%=mod;}}}}int t;cin>>t;while(t--){int n,m;cin>>n>>m;LL result=0;for(int i=1;i<=m;i++)result=(result+f[n][m][i])%mod;cout<<result<<endl;}return 0;}

D

E

#include<iostream>using namespace std;const int maxn=200000+5;char s[maxn];int main(){int t;scanf("%d",&t);while(t--){int n;scanf("%d",&n);scanf("%s",s);int len=0;long long count=0;for(int i=0;i<n;i++){if(s[i]=='6') len++;else{count+=(long long)(len+1)*len/2;len=0;}}if(len) count+=(long long)(len+1)*len/2;printf("%lld\n",count);}return 0;}

F

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H

#include<iostream>using namespace std;typedef long long LL;const int mod=1e9+7;struct Matrix{int m[2][2];}base,ans;Matrix multi(Matrix a,Matrix b){Matrix tmp;for(int i=0;i<2;i++){for(int j=0;j<2;j++){tmp.m[i][j]=0;for(int k=0;k<2;k++){tmp.m[i][j]=(tmp.m[i][j]+1LL*a.m[i][k]*b.m[k][j])%mod;}}}return tmp;}LL Fibonacci(LL n){ans.m[0][0]=ans.m[1][1]=1;ans.m[0][1]=ans.m[1][0]=0;base.m[0][0]=base.m[0][1]=base.m[1][0]=1;base.m[1][1]=0;while(n){if(n&1)ans=multi(ans,base);base=multi(base,base);n>>=1;}return ans.m[0][0];}int main(){LL n;while(scanf("%lld",&n)&&n){printf("%lld\n",Fibonacci(n));}return 0;}

I

#include<iostream>#include<algorithm>using namespace std;typedef long long LL;const int maxn=100000+5;LL num[maxn];int main(){int n;LL k;while(~scanf("%d%lld",&n,&k)){for(int i=0;i<n;i++) scanf("%lld",&num[i]);sort(num,num+n);int pos=0,count=0;;for(int i=0;i<n;i++){LL tmp=num[i]+k;while(num[pos]<tmp&&pos<n) pos++;if(num[pos]==tmp){int pos1=pos;while(num[pos1]==tmp&&pos<n) {pos1++;count++;}}}printf("%d\n",count);}return 0;}

J

#include<iostream>using namespace std;int main(){int n;while(~scanf("%d",&n)){int x,count=0;for(int i=0;i<n;i++) {scanf("%d",&x);if(x%2==1) count++;}printf("%d\n",count);}return 0;}

K

#include<iostream>using namespace std;const int mod=1e9+7;int pow_mod(int a,int b){int result=1;int base=a;while(b){if(b&1) result=(1LL*result%mod*base%mod)%mod;base=(1LL*base*base%mod%mod)%mod;b>>=1;}return result;}int main(){int t;cin>>t;while(t--){int n,m;cin>>n>>m;cout<<pow_mod(m,n)<<endl;}return 0;}

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