HDU-#4813 Hard Code

本文介绍了一道编程题的解决方法,题目要求将输入的字符串按指定的行数和列数输出成矩阵形式。通过简单的循环结构实现,适用于初学者练习。

        题目大意:给出一个字符串以及n,m,要求将其按照n行m列进行输出。

       解题思路:直接就暴力输出,按照先行后列进行输出即可。其实更快的写法是边接收边输出。

       题目来源:http://acm.hdu.edu.cn/showproblem.php?pid=4813

      code:

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

const int MAXN = 1000+10;
int l,m,n;
char str[MAXN];

int main(){
    //freopen("input.txt","r",stdin);
    scanf("%d",&l);
    while(l--){
        memset(str,0,sizeof(str));
        scanf("%d%d",&n,&m);
        scanf("%s",str);
        for(int i=0;i<n;++i){
            int tmp=i*m;
            for(int j=0;j<m;++j){
                printf("%c",str[tmp+j]);
            }
            printf("\n");
        }
    }
    return 0;
}

现在不会TLE了,但还是WA了: #include <bits/stdc++.h> using namespace std; #define int long long const int INF = 0x3f3f3f3f; const int MAX_N = 2e5 + 50; int T, n, m, a[MAX_N]; #define ls(cur) cur << 1 #define rs(cur) cur << 1 | 1 namespace SegmentTree1 { // the case of a[i] <= x long long sum[MAX_N << 2], vtag[MAX_N << 2]; int cnt[MAX_N << 2], stag[MAX_N << 2]; void pushup(int cur) { sum[cur] = sum[ls(cur)] + sum[rs(cur)]; cnt[cur] = cnt[ls(cur)] + cnt[rs(cur)]; } void mark(int cur, int sign, long long val) { sum[cur] *= sign; sum[cur] += cnt[cur] * val; stag[cur] *= sign; vtag[cur] += val; } void pushdown(int cur) { if (stag[cur] != 1) { mark(ls(cur), stag[cur], 0); mark(rs(cur), stag[cur], 0); stag[cur] = 1; } if (vtag[cur]) { mark(ls(cur), 1, vtag[cur]); mark(rs(cur), 1, vtag[cur]); vtag[cur] = 0; } } void build(int cur, int l, int r) { stag[cur] = 1; vtag[cur] = 0; if (l == r) { sum[cur] = cnt[cur] = 0; return ; } int mid = l + r >> 1; build(ls(cur), l, mid); build(rs(cur), mid + 1, r); pushup(cur); } void insert(int cur, int l, int r, int idx, int val) { if (l == r) { sum[cur] = val; cnt[cur] = 1; return ; } pushdown(cur); int mid = l + r >> 1; if (idx <= mid) insert(ls(cur), l, mid, idx, val); else insert(rs(cur), mid + 1, r, idx, val); pushup(cur); } void modify(int cur, int l, int r, int L, int R, int val) { if (L <= l && r <= R) { mark(cur, -1, val); return ; } pushdown(cur); int mid = l + r >> 1; if (L <= mid) modify(ls(cur), l, mid, L, R, val); if (mid + 1 <= R) modify(rs(cur), mid + 1, r, L, R, val); pushup(cur); } long long query(int cur, int l, int r, int L, int R) { if (L <= l && r <= R) return sum[cur]; pushdown(cur); int mid = l + r >> 1; long long res = 0; if (L <= mid) res += query(ls(cur), l, mid, L, R); if (mid + 1 <= R) res += query(rs(cur), mid + 1, r, L, R); return res; } }; namespace SegmentTree2 { // the case of a[i] > x long long sum[MAX_N << 2], tag[MAX_N << 2]; int cnt[MAX_N << 2], mn[MAX_N << 2]; void pushup(int cur) { sum[cur] = sum[ls(cur)] + sum[rs(cur)]; cnt[cur] = cnt[ls(cur)] + cnt[rs(cur)]; mn[cur] = min(mn[ls(cur)], mn[rs(cur)]); } void mark(int cur, long long val) { sum[cur] -= cnt[cur] * val; mn[cur] -= bool(cnt[cur]) * val; tag[cur] += val; } void pushdown(int cur) { if (tag[cur]) { mark(ls(cur), tag[cur]); mark(rs(cur), tag[cur]); tag[cur] = 0; } } void build(int cur, int l, int r, int val[]) { tag[cur] = 0; if (l == r) { sum[cur] = mn[cur] = val[l]; cnt[cur] = 1; return ; } int mid = l + r >> 1; build(ls(cur), l, mid, val); build(rs(cur), mid + 1, r, val); pushup(cur); } void modify(int cur, int l, int r, int L, int R, int val) { if (l == r && mn[cur] <= val) { SegmentTree1::insert(1, 1, n, l, mn[cur]); sum[cur] = cnt[cur] = 0, mn[cur] = INF; return ; } if (L <= l && r <= R && mn[cur] > val) { mark(cur, val); return ; } pushdown(cur); int mid = l + r >> 1; if (L <= mid) modify(ls(cur), l, mid, L, R, val); if (mid + 1 <= R) modify(rs(cur), mid + 1, r, L, R, val); pushup(cur); } long long query(int cur, int l, int r, int L, int R) { if (L <= l && r <= R) return sum[cur]; pushdown(cur); int mid = l + r >> 1; long long res = 0; if (L <= mid) res += query(ls(cur), l, mid, L, R); if (mid + 1 <= R) res += query(rs(cur), mid + 1, r, L, R); return res; } }; signed main() { ios::sync_with_stdio(false); cin.tie(0), cout.tie(0); cin >> T; while (T--) { cin >> n >> m; for (int i = 1; i <= n; i++) cin >> a[i]; SegmentTree1::build(1, 1, n); SegmentTree2::build(1, 1, n, a); for (int opt, l, r, x; m--; ) { cin >> opt >> l >> r; if (opt == 1) { cin >> x; SegmentTree2::modify(1, 1, n, l, r, x); SegmentTree1::modify(1, 1, n, l, r, x); } else { long long ans1 = SegmentTree1::query(1, 1, n, l, r); long long ans2 = SegmentTree2::query(1, 1, n, l, r); cout << ans1 + ans2 << &#39;\n&#39;; } } } return 0; }
07-29
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