2020初中生暑假 第一套

t1扫雷

二维数组，方向数组

#include<cstdio>
using namespace std;
char c[105][105];int a[105][105],n,m;
int dx[]={0,1,1,1,0,-1,-1,-1},dy[]={1,1,0,-1,-1,-1,0,1};
int main(){
    freopen("game.in","r",stdin);
    freopen("game.out","w",stdout);
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;++i){
        scanf("%s",c[i]+1);
        for(int j=1;j<=m;++j)
            if(c[i][j]=='#')
                for(int k=0;k<8;++k)
                    ++a[i+dx[k]][j+dy[k]];
    }
    for(int i=1;i<=n;++i){
        for(int j=1;j<=m;++j)
            c[i][j]=='#'?printf("#"):printf("%d",a[i][j]);
        printf("\n");
    }
    return 0;
} 

 

t2 music

枚举每一段旋律，即单词，对应枚举歌曲的每一个位置，判断是否一一对应相同，相同则累加分数。

#include<cstdio>
#include<cstring>
using namespace std;
char c[15][105],s[105];int n,w[15],l,ans;
bool check(int x,int len,int wz){
    for(int i=0;i<len;++i)
        if(c[x][i]!=s[i+wz])
            return 0;
    return 1;
}
int main(){
    // freopen("music.in","r",stdin);
    // freopen("music.out","w",stdout);
    scanf("%d",&n);
    for(int i=1;i<=n;++i)
        scanf("%s%d",c[i],&w[i]);
    scanf("%s",s);
    l=strlen(s);
    for(int i=1;i<=n;++i){
        int l1=strlen(c[i]);
        for(int j=0;j<=l-l1;++j)
            if(check(i,l1,j)) ans+=w[i];
    }
    printf("%d",ans);
    return 0;
}

 t3 sum 最大子段和

//这是一道很好的最大子段和的题目，正着一遍，倒着一遍
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<queue>
#define int long long
using namespace std;
int n,f[100005],g[100005],a[100005],ans = -0x7fffffff,mxf[100005],mxg[100005];
signed main(){
    scanf("%lld",&n);
    for(int i = 1; i <= n; i++){
        scanf("%lld",&a[i]);
    }
    //以当前位置结尾的最大子段和,同时记录在它之前的最大子段和的值
    for(int i = 1; i <= n; i++){
        if(f[i-1] < 0) f[i] = a[i];
        else f[i] = f[i-1] + a[i];
        ans = max(ans, f[i]);
        mxf[i] = ans;
        // cout<<mxf[i]<<" ";
    }

    //倒着来一遍
    ans = -0x7fffffff;
    for(int i = n; i >= 1; i--){
        if(g[i+1] < 0) g[i] = a[i];
        else g[i] = g[i+1] + a[i];
        ans = max(ans, g[i]);
        mxg[i] = ans;
    }

    //枚举分界点
    for(int i = 1; i <= n; i++)
        if(mxf[i] + mxg[i+1] > ans) ans = mxf[i]+mxg[i+1];
    printf("%lld\n",ans);
    return 0;
}
/*
7
2 -4 3 -1 2 -4 3
*/